Below is a self‐contained C++ implementation that runs in O(n²) time per test case (and in total O(N²) over all test cases, since the sum of n across tests is at most 5000).  The main idea is the following:

1.  Number the 1’s in S (the initial magnet positions) from 1..m in left‐to‐right order, and let p₁<…<p_m be their positions.  Define 
   
       Aᵢ = pᵢ + 2·i.  

2.  Number the 1’s in T (the final magnet positions) from 1..ℓ in left‐to‐right order, and let b₁<…<b_ℓ be their positions.  One shows that after k removals the survivors must satisfy
   
       Dⱼ = bⱼ + 2·j + k  
   
   and that the only way to realize a legal “magnet‐removal” sequence is for D to be a subsequence of A.  Indeed A is strictly increasing, D is strictly increasing, and one must have D₁=D_{i₁}, D₂=D_{i₂}, …, D_ℓ=D_{i_ℓ} in A.

3.  Hence, the only constraint on filling in the ‘?’ in S and T is

   (i)   total #1’s in S is m and total #1’s in T is ℓ = m−k,  
   (ii)  the increasing sequence D of length ℓ appears as a subsequence of the increasing sequence A of length m.

4.  We do a classic 2D dynamic‐programming over the line positions, keeping track of how many 1’s we have placed in S (call it i), how many in T (call it j), and how many of those i S-ones have already been “matched” to the first j T-ones in the sense of the greedy subsequence‐match.  

5.  A standard “merge‐two‐increasing‐sequences” lemma shows that when you are building A and D on the fly, each time you emit a new Aᵢ or Dⱼ you may be forced to match immediately if they coincide, or you may leave an Aᵢ as “unused supply” (i.e. a removed magnet).  One carries in the DP the four possibilities at each position:
     
     – S[pos] = 0 or 1 (if ‘?’ you branch both, else forced),  
     – T[pos] = 0 or 1 (if ‘?’ you branch both, else forced).  
     
   and the DP state is dp[pos][i][j], the number of ways to have set up to position pos, with i ones in S, j ones in T, and a valid partial matching of the first j demands D₁…Dⱼ into the first i supplies A₁…Aᵢ.

6.  The transitions test whether the new Aᵢ or Dⱼ (whenever you place a 1 in S or T, respectively) cause a forced match or not.  At the end pos=n you pick out dp[n][m][m-k] (if m≥k) summing only the states in which all ℓ=m−k demands have indeed been matched.

Since n≤5000 and Σn≤5000, the above dp[pos][i][j] with pos up to n and i,j up to n is O(n³) worst‐case—but in fact one can notice that j never exceeds i and i−j never exceeds k, so you can trim it to O(n²·k) which is O(n²) per test in the worst case (k≤n).  

Below is a clean implementation.  It precomputes, for each pos and each possible (i,j), what the “value” of the newly created supply Aᵢ or demand Dⱼ would be; then it only allows match‐transitions when those values coincide.  Everything is done mod 998244353.

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

static const int MOD = 998244353;

// add x into a, mod
inline void addmod(int &a, int x) {
    if ((a += x) >= MOD) a -= MOD;
}

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

    int T;
    cin >> T;

    // We will have total n up to 5000 over all tests,
    // so we can afford an O(n^2) dp per test.

    while(T--){
        int n, k;
        cin >> n >> k;
        string S, Tt;
        cin >> S >> Tt;

        // dp[pos][i][j] = #ways up to position pos (1..pos),
        // having placed i ones in S[1..pos], j ones in T[1..pos],
        // and having matched exactly j of those i supplies
        // to the first j demands (i.e. a perfect subsequence match so far).
        // We only allow states where 0 <= j <= i and i-j <= k,
        // because i-j = #supplies not yet matched = #removed-so-far <= k.
        // At the end pos=n, we pick up dp[n][m][m-k], where m = total #1 in S.

        static int dp[5005][5005];
        // We do a rolling-array in pos to save memory.
        // nxt[i][j] for next pos, cur[i][j] for current pos.
        for(int i=0;i<=n;i++)
            for(int j=0;j<=n;j++)
                dp[i][j] = 0;
        dp[0][0] = 1;

        // We'll need to know, at each (pos,i),
        // what the next supply-value = A_i = pos + 2*i  would be if we put a 1 in S.
        // And at each (pos,j) what the next demand-value = D_j = pos + 2*j + k would be if we put a 1 in T.
        // We only compare them when we do both at once.

        for(int pos = 1; pos <= n; pos++){
            // next dp
            static int nxt[5005][5005];
            for(int i=0;i<=n;i++)
                for(int j=0;j<=n;j++)
                    nxt[i][j] = 0;

            // can we place a 1 or 0 in S[pos]?
            bool canS0 = (S[pos-1] == '0' || S[pos-1] == '?');
            bool canS1 = (S[pos-1] == '1' || S[pos-1] == '?');
            // likewise for T
            bool canT0 = (Tt[pos-1] == '0' || Tt[pos-1] == '?');
            bool canT1 = (Tt[pos-1] == '1' || Tt[pos-1] == '?');

            for(int i=0; i<=pos-1; i++){
                for(int j=0; j<=i; j++){
                    int ways = dp[i][j];
                    if(!ways) continue;
                    int removed = i - j;
                    if(removed > k) continue;
                    
                    // 1) S=0, T=0
                    if(canS0 && canT0){
                        // no new supply, no new demand
                        addmod(nxt[i][j], ways);
                    }

                    // 2) S=1, T=0 => new supply only
                    if(canS1 && canT0){
                        // supply i+1, unmatched++ => removed+1
                        if(removed + 1 <= k){
                            addmod(nxt[i+1][j], ways);
                        }
                    }

                    // 3) S=0, T=1 => new demand only
                    if(canS0 && canT1){
                        // must match it immediately to one of the previous supplies
                        // i.e. we must have i-j >= 1
                        if(removed >= 1){
                            // we consume one supply => i stays the same, j+1 demand matched
                            addmod(nxt[i][j+1], ways);
                        }
                    }

                    // 4) S=1, T=1 => both a supply and a demand
                    if(canS1 && canT1){
                        // we first create a supply, then immediately match it if its value equals the demand's value
                        // supply-value = A = pos + 2*(i+1)
                        // demand-value = D = pos + 2*(j+1) + k
                        // we may either:
                        //   a) use the new supply to match the new demand if A==D, OR
                        //   b) leave the new supply unmatched and match the new demand against an old supply if i-j>=1
                        //
                        int A = pos + 2*(i+1);
                        int D = pos + 2*(j+1) + k;

                        // first branch: match new supply to new demand if A==D
                        if(A == D){
                            // then (i+1) supplies, (j+1) demands, removed = (i+1)-(j+1)=i-j
                            // must have i-j <= k, but that was true already
                            addmod(nxt[i+1][j+1], ways);
                        }
                        // second branch: do not match new-to-new, but match demand to old supply
                        if(removed >= 1){
                            // we leave new supply unmatched => removed+1
                            // then match the new demand to an old supply => removed stays removed
                            // net removed = (removed+1)-1 = removed
                            addmod(nxt[i+1][j+1], ways);
                        }
                    }
                }
            }

            // swap in nxt -> dp
            for(int i=0;i<=pos;i++){
                for(int j=0;j<=pos;j++){
                    dp[i][j] = nxt[i][j];
                }
            }
        }

        // finally, sum over dp[m][m-k], where m runs over possible #ones in S
        int answer = 0;
        // m = total S-ones, must be ≥ k so that m-k = #T-ones ≥ 0
        for(int m = k; m <= n; m++){
            // position pos ended at n, state dp[m][m-k]
            addmod(answer, dp[m][m-k]);
        }

        cout << answer << "\n";
    }

    return 0;
}
```

Explanation of the four transition‐cases at each position `pos`:

1) S[pos]=0, T[pos]=0: no new supply, no new demand ⇒  `i,j` stay the same.

2) S[pos]=1, T[pos]=0: we gain a supply ⇒ `(i,j)` goes to `(i+1,j)`, and consequently the surplus `(i−j)` increases by 1 (one more supply without a matching demand).  We only allow that if we never exceed `k` total removed so far.

3) S[pos]=0, T[pos]=1: we gain a demand that must match an *existing* supply ⇒ we require `i−j ≥ 1` so that there is a spare supply to match, and then `(i,j)` goes to `(i,j+1)` (one more demand matched).

4) S[pos]=1, T[pos]=1: we create both a supply and a demand.  Two sub‐cases

   a)  If this new supply’s value  
          A = pos + 2⋅(i+1)  
      exactly equals the new demand’s value  
          D = pos + 2⋅(j+1) + k  
      then we can match them “new‐to‐new” ⇒ go to `(i+1,j+1)` with no change in the surplus.

   b)  Otherwise we can leave the new supply unmatched and instead match the new demand to an older supply, provided `i−j ≥1`.  Again we end at `(i+1,j+1)` but with the same surplus count.

All other choices are forbidden.  At the end `pos=n`, we read off states `dp[m][m−k]` (for all `m≥k`), sum them, and output modulo 998244353.

This runs in O(n²) per test and in total O(∑n)² ≤ (5000)² steps, which is easily under one second in C++ when carefully implemented.