**Analysis**

We have up to 52 distinct ordinary playing‐cards of the four suits D, C, H, S (no wilds initially), and up to one copy of each of six “Tarot” cards which can be used at most once each (or not at all).  The four “elemental” cards (Star, Moon, Sun, World) can each convert *up to* 3 cards into diamonds, clubs, hearts, or spades respectively; The Lovers can convert exactly 1 card into a wild; Death can copy one chosen card onto another chosen card (so effectively takes one card of some suit or wild and turns a second card into an extra copy).

After all our chosen uses of those special powers, we end up with

  - some final counts \(c_D,c_C,c_H,c_S\) of ordinary cards of each suit,
  - and \(w\) wild cards,

with total \(c_D+c_C+c_H+c_S + w = n\) (the hand size never changes).  A *flush* is any 5 cards all of which are either of the same suit \(X\) or wild; once you peel off such a 5–card flush you discard them and cannot reuse them.

We must choose how many conversions of each type to do (up to the given limits), possibly use The Lovers (to create up to 1 wild) and/or Death (which can give one extra “unit” either in one suit or in the wild pile), so as to **maximize** the number of disjoint 5‐card flushes we can peel off.

Key observations:

1.  Since *which* cards of a suit or which exact ranks we convert is immaterial to flush‐counting (only the *counts* of each suit and of wild matter), we can boil every choice down to **how many cards** we net convert into each suit and into wild.

2.  If we decide to convert \(D_D\) cards into diamonds, \(D_C\) into clubs, \(D_H\) into hearts, \(D_S\) into spades, and \(D_w\) into wild, then that creates
   \[
     c^\text{tot}_X = c^0_X \;+\; D_X 
     \quad(X\in\{D,C,H,S\}),\quad
     w = D_w,
   \]
   where \(c^0_X\) is the original count.  But **every** conversion must pick a card out of your hand and overwrite it; thus you must “spend” exactly
   \[
     T \;=\; D_D+D_C+D_H+D_S + D_w
   \]
   cards total as the *sources* of those conversions.  You may choose your \(T\) source‐cards from any suits (wilds cannot be used as sources for suit‐conversions, but we will only source out of ordinary suits).  Converting a diamond into a diamond still counts as “using up” one of your diamonds if you run out of any other suit; in effect, you may take your \(T\) source‐cards *in any distribution* from the four suits, as long as you do not exceed your original \(c^0_X\) in suit \(X\).  

   Hence the only global feasibility check is 
   \[
     T \;\le\; c^0_D + c^0_C + c^0_H + c^0_S \;=\; n.
   \]
   (No more subtle per‐suit supply constraints arise, because you *can* choose same‐suit conversions if you must.)

3.  Once you have
   \[
     c^\text{tot}_X = c^0_X + D_X,
     \quad
     w = D_w,
     \quad
     T = \sum D,
   \]
   you must pick exactly \(T\) cards *out of* your originals as the sources.  Obviously you will pick them so as *least* damage your ultimate ability to form flushes.  In particular you will discard from the suits that have the smallest \(c^\text{tot}_X\) first.  

   Concretely, you do:

   (a) form a list of the four suits sorted by  
         \(\,c^\text{tot}_X\) ascending,  
   (b) withdraw up to your remaining budget \(T\) from the smallest suit’s *original* cards, then the next smallest, etc.  

   That determines your final
   \[
     c_X = c^\text{tot}_X \;-\;\text{(how many you withdrew from suit }X),
     \quad w \text{ unchanged.}
   \]

4.  Finally, you solve a small integer‐DP or greedy check to see how many 5‐card flushes can be carved out of \((c_D,c_C,c_H,c_S;w)\).  

Because

  - Each “elemental” can give up to 3 conversions → at most 3 each of D, C, H, S,
  - Lovers can give up to 1 wild,
  - Death (if used) gives exactly +1 to *one* of D/C/H/S or to the wild pile,

there are only a few thousand total choices of \((D_D,D_C,D_H,D_S,D_w)\) to try.  For each one we

  1.  Check that the net number \(T\) of sources we need does not exceed \(n\).
  2.  “Greedily withdraw” those \(T\) sources from the suits in ascending order of their intermediate counts.
  3.  Run a tiny DP to find the maximum number of flushes from the resulting \((c_D,c_C,c_H,c_S;w)\).

Overall complexity per test is on the order of a few thousand × a few hundred steps, which easily fits under 1 second even for up to \(T\le13\) test cases.

---

```cpp
#include <bits/stdc++.h>
using namespace std;

/*
 We map suits D, C, H, S to indices 0,1,2,3.  We also track "wild" as index 4 in our D_* arrays.
 
 c0[s] = original count of suit s in {0..3}.  No wilds originally.
 t1,t2,t3,t4 = whether we have Star, Moon, Sun, World (each 0/1).
 t5 = Lovers, t6 = Death.
 
 cap_op[s] = how many net conversions we can do into suit s from the 4 elemental cards:
   cap_op[0] = 3*t1  (Star → Diamonds)
   cap_op[1] = 3*t2  (Moon → Clubs)
   cap_op[2] = 3*t3  (Sun → Hearts)
   cap_op[3] = 3*t4  (World → Spades)
   cap_op[4] = t5    (Lovers → wild)
 
 If we do *not* use Death at all, then D_s ∈ [0..cap_op[s]] for s=0..4.
 If we *do* use Death exactly once, it gives one extra +1 to exactly one of
    {D, C, H, S, wild}.  So we loop "extra=0..t6", and if extra=1 we decide
    which index X∈[0..4] gets +1 cap.
 
 For each choice (extra, & if extra=1 then target X), we enumerate
    D_s ∈ [0 .. cap_op[s] + (extra==1 && s==X ? 1 : 0)]
 but we require D_X ≥ 1 if extra=1 (otherwise we didn't actually use Death).
 
 Let T = sum(D_s).  If T > n (total cards) skip it.
 Otherwise we have intermediate counts
    c_tot[s] = c0[s] + D_s   for s=0..3
    w        = D_4          (the wild count)
 We must "withdraw" T cards from the union of suits 0..3 so as to harm us least.
 We greedily remove first from the suit with smallest c_tot[s], then next, ...
   withdraw[s] = min(c0[s], remaining T) in that ascending order.
 Then
   c_final[s] = c_tot[s] - withdraw[s],
   w_final    = w.
 
 Finally we do a small DP to see how many 5-card flushes we can form from
   (c_final[0..3], w_final).
 We keep a best‐so-far over all these enumerations.
*/

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int T; 
    cin >> T;
    while(T--){
        int n;
        cin >> n;
        vector<int> c0(4,0);
        for(int i=0;i<n;i++){
            string card;
            cin >> card;
            char s = card[1];
            int idx = (s=='D'?0:(s=='C'?1:(s=='H'?2:3)));
            c0[idx]++;
        }
        int t1,t2,t3,t4,t5,t6;
        cin >> t1 >> t2 >> t3 >> t4 >> t5 >> t6;

        // cap_op[s] = how many suit‐conversions we get from the 4 elementals,
        // and cap_op[4] = how many wilds from Lovers.
        array<int,5> baseCap;
        baseCap[0] = 3*t1;    // Diamonds from Star
        baseCap[1] = 3*t2;    // Clubs    from Moon
        baseCap[2] = 3*t3;    // Hearts   from Sun
        baseCap[3] = 3*t4;    // Spades   from World
        baseCap[4] = t5;      // wild     from Lovers

        int best = 0;

        // We will try extra=0 (no Death used) and if t6=1 also extra=1 (Death used once),
        // distributing that +1 to exactly one of the five targets.
        for(int extra=0; extra<=t6; extra++){
            // We'll pick a "deathTarget" in 0..4 if extra=1,
            // or no target if extra=0.
            for(int deathTarget = (extra==0? -1 : 0);
                    deathTarget < (extra==0? 0:5);
                    deathTarget++)
            {
                // Build the caps for D_s.
                array<int,5> cap = baseCap;
                if(extra==1){
                    cap[deathTarget]++;
                }

                // Now enumerate D_s in [0..cap[s]].
                // If extra=1 we demand D[deathTarget]>=1 else we really didn't use Death.
                for(int dD = 0; dD <= cap[0]; dD++){
                 for(int dC = 0; dC <= cap[1]; dC++){
                  for(int dH = 0; dH <= cap[2]; dH++){
                   for(int dS = 0; dS <= cap[3]; dS++){
                    for(int dw = 0; dw <= cap[4]; dw++){
                        if(extra==1){
                            int v = (deathTarget==0? dD
                                     : deathTarget==1? dC
                                     : deathTarget==2? dH
                                     : deathTarget==3? dS
                                     : dw);
                            if(v<1) continue;  // we said we'd use Death → must see +1 used
                        }
                        int Dsum = dD + dC + dH + dS + dw;
                        if(Dsum > n) continue;  // cannot withdraw that many cards

                        // intermediate totals after conversions:
                        array<int,4> ctot;
                        ctot[0] = c0[0] + dD;
                        ctot[1] = c0[1] + dC;
                        ctot[2] = c0[2] + dH;
                        ctot[3] = c0[3] + dS;
                        int wfin = dw;

                        // Withdraw Dsum cards, greedily from the suits with smallest ctot[s].
                        array<pair<int,int>,4> order;
                        for(int s=0;s<4;s++){
                            order[s] = { ctot[s], s };
                        }
                        sort(order.begin(), order.end());
                        array<int,4> withdraw{0,0,0,0};
                        int rem = Dsum;
                        for(int i=0;i<4 && rem>0;i++){
                            int s = order[i].second;
                            int take = min(rem, c0[s]);
                            withdraw[s] = take;
                            rem -= take;
                        }
                        if(rem>0) {
                            // means we didn't have enough original cards to withdraw,
                            // but we already checked Dsum<=n so this won't happen.
                            continue;
                        }
                        array<int,4> cfin;
                        for(int s=0;s<4;s++){
                            cfin[s] = ctot[s] - withdraw[s];
                        }

                        // Now compute how many flushes we can make from (cfin, wfin).
                        // We'll DP over suits, distributing kD,kC,kH,kS flushes.
                        // Max possible K <= (sum cfin + wfin)/5 <= n/5.
                        int maxK = (cfin[0]+cfin[1]+cfin[2]+cfin[3] + wfin)/5;
                        if(maxK <= best) continue; // no improvement possible
                        int localBest = best;

                        // Try K = maxK down to best+1, break as soon as feasible.
                        // DP[i][k][w] = can we use first i suits to make k flushes
                        // consuming exactly w wilds?
                        for(int K = maxK; K>best; K--){
                            static bool dp[5][16][53];
                            memset(dp,false,sizeof(dp));
                            dp[0][0][0] = true;
                            for(int i=0;i<4;i++){
                                for(int k=0;k<=K;k++){
                                    for(int wU=0;wU<=wfin;wU++){
                                        if(!dp[i][k][wU]) continue;
                                        // try x= how many flushes we assign to suit i
                                        // max x <= K-k, and also 5*x <= cfin[i]+wfin-wU
                                        int capx = (cfin[i]+(wfin-wU))/5;
                                        capx = min(capx, K-k);
                                        for(int x=0;x<=capx;x++){
                                            // wild needed = max(0, 5*x - cfin[i])
                                            int need = max(0,5*x - cfin[i]);
                                            if(wU+need <= wfin){
                                                dp[i+1][k+x][wU+need] = true;
                                            }
                                        }
                                    }
                                }
                            }
                            bool ok = false;
                            for(int wU=0;wU<=wfin;wU++){
                                if(dp[4][K][wU]){
                                    ok = true;
                                    break;
                                }
                            }
                            if(ok){
                                localBest = K;
                                break;
                            }
                        }
                        best = max(best, localBest);

                    }}}}
                if(extra==0) break; // only one iteration for extra=0
            }
        }

        cout << best << "\n";
    }
    return 0;
}
```

**Explanation of the Main Steps**

1. **Count the original suits**:  
   We read the \(n\) cards and tally how many Diamonds, Clubs, Hearts, Spades we have initially.  There are no wilds to begin with.

2. **Set up conversion‐limits**  
   - `cap_op[0] = 3*t1` for Diamonds (Star),  
     `cap_op[1] = 3*t2` for Clubs (Moon),  
     `cap_op[2] = 3*t3` for Hearts (Sun),  
     `cap_op[3] = 3*t4` for Spades (World),  
     `cap_op[4] = t5`   for wilds (Lovers).

3. **Enumerate whether we use Death**  
   - `extra=0` means “do not use Death.”  
   - `extra=1` (only if `t6=1`) means “use Death exactly once,”  
     and we must choose which of the 5 “buckets” (the 4 suits or the wilds) gets that extra +1.  
   That simply allows one extra unit beyond the base caps.

4. **Enumerate how many cards we net convert into each of the 5 buckets**  
   We do 5‐nested loops for  
      \(D_D\in[0..\mathrm{cap}_D],\,D_C\in[0..\mathrm{cap}_C],\dots,D_w\in[0..\mathrm{cap}_w]\).  
   If `extra=1`, we additionally require that the chosen “death‐bucket” actually get at least 1 so as to guarantee we did indeed *use* Death.

5. **Check feasibility of “withdrawing” source cards**  
   Converting a total of \(T=D_D+D_C+D_H+D_S+D_w\) cards requires picking \(T\) source cards to overwrite.  Since we only have \(n\) cards, we must have
   \[
     T \;\le\; n.
   \]
   If not we skip.

6. **Compute intermediate totals**  
   \[
     \text{ctot}[s] = c^0_s + D_s
     \quad(s=0..3),\qquad
     w = D_w.
   \]

7. **Greedy withdrawal**  
   We must remove \emph{some} \(T\) cards from our 4 suits as the sources of those conversions.  We do this from
   the suits with smallest `ctot[s]` first (because those suits will contribute least to making flushes).  
   If a suit has only \(c^0_s\) original cards, we can remove at most that many from it (we cannot remove more originals than we had).

   After we withdraw exactly \(T\), we get our final
   \[
     c_s = \text{ctot}[s] \;-\;\text{withdrawn}[s],
     \quad
     w \;=\; D_w.
   \]

8. **DP to count maximum flushes**  
   With \((c_D,c_C,c_H,c_S;w)\) in hand, we ask: “How many disjoint 5-card flushes can I carve off?”  
   We do a small DP over how many flushes we assign to each suit, keeping track of how many wilds we must spend.  
   The maximum over all suits \(\le 10\) flushes is found quickly.

9. **Take the best over all enumerations**.  

This runs in a few thousand enumerations × a small‐constant DP each, easily under 1 s for all test‐cases.
