**Analysis**

We have two binary‐string states of length $n$, an initial state $S$ (with “?” wildcards) and a final state $T$ (also with “?”), and a fixed integer $k$.  A “legal experiment” consists of

 1. Starting with the magnets in the positions of the 1’s of $S$.
 2. Repeating exactly $k$ times:
    • Pick one of the *current* magnets, remove it,  
    • all remaining magnets strictly *left* of it shift 1 unit left,  
    • all remaining magnets strictly *right* of it shift 1 unit right.
 3. At the end, the magnets lie exactly in the positions specified by the 1’s of $T$ (all between 1 and $n$).

We must count, over all ways to replace the “?”’s in $S$ and $T$ by 0/1, how many resulting pairs $(S,T)$ admit at least one legal experiment.  Finally we return the count **modulo** $998244353$.

---

### 1.  The key “shift”‐formula

Label the ones of $S$ in *ascending* coordinate order: suppose there are $m$ total ones, at positions
$$
   p_1<p_2<\cdots<p_m.
$$
When we remove $k$ of those magnets one‐by‐one, exactly $m-k$ survive; let their *original* indices among $p_\ast$ be
$$
   i_1<i_2<\cdots<i_{\,m-k}.
$$
It is a standard “shifting” argument to show that the $u$–th *survivor* (i.e.\ the one originally at coordinate $p_{i_u}$) ends up at final coordinate
$$
   p_{\,i_u}\;+\;2\,\bigl(\#\hbox{removed among indices }<i_u\bigr)\;-\;k.
$$
Since among the first $i_u$ magnets we have removed exactly 
$$
   L_u \;=\;\#\{\;\text{removed among }1,2,\dots,i_u\}\,,
$$ 
the final coordinate of that survivor is
$$
   p_{\,i_u}+2L_u - k.
$$

On the other hand the *target* final magnets are exactly the 1’s of $T$, say there are $m'=\,m-k$ ones at positions
$$
   q_1<q_2<\cdots<q_{\,m'}.
$$
They must be matched in order, so the $u$–th survivor must land at $q_u$.  Hence for each $u=1,2,\dots,m'$ we require
$$
   p_{\,i_u} \;+\;2\,L_u \;-\;k\;=\;q_u,
$$
where 
$$
   L_u \;=\; i_u - u
$$
(since among the first $i_u$ magnets, exactly $u$ of them survive, so the other $i_u-u$ were removed).  Plugging in $L_u=i_u-u$ gives the single‐equation

    p_{\,i_u} \;+\;2\,(i_u - u)\;-\;k \;=\;q_u.
    
Rearrange it as
          
    q_u \;=\; p_{\,i_u} \;+\;2\,i_u\;-\;2u\;-\;k.
    
---

### 2.  Uniqueness and “subsequence” criterion

Two facts are crucial:

 1.  If the initial positions $p_i$ satisfy $p_{i+1}\ge p_i+1$, then the quantities
    $$
       C_i \;=\; p_i \;+\;2\,i \;-\;k
    $$
    form a strictly increasing sequence (by at least 3 each time).    
 2.  Likewise the target quantities
    $$
       D_u \;=\; q_u \;+\;2\,u
    $$
    form a strictly increasing sequence.

Matching the $u$–th survivor means finding $i_u$ so that
$$
   C_{\,i_u}\;=\;D_u.
$$
Because both $\{C_i\}$ and $\{D_u\}$ are strictly increasing, if a matching exists it is *unique*: for each $u$ there is at most one $i$ with $C_i=D_u$, and those $i$ must rise as $u$ does.  Hence the count of ways to choose *which* $k$ we remove is either $0$ or $1$ once the two bit‐patterns are fully specified.

Therefore **the only remaining question**, once $S$ is fixed (so $p_i$ are known) and $T$ is fixed (so $q_u$ are known), is:

    “Is the sequence $D_1,D_2,\dots,D_{m'}$ a subsequence of $C_1,C_2,\dots,C_m$?”

If **yes**, there is exactly one way to remove the $k$ “extra” magnets so that you land on exactly $T$.  If **no**, there are zero ways.

---

### 3.  Bringing in the “?” wildcards

We now must sum (mod $998244353$) over all completions of the “?”’s in $S$ and $T$ of the indicator of that subsequence‐condition.

A direct double‐sum over $2^{\#?_S}\times2^{\#?_T}$ is impossible.  The classic trick is to do a 2D DP over the *positions* $x=0,1,2,\dots,n$ and keep track of how many ones we have so far in $S$ and in $T$, and maintain the progress of merging the two strictly‐increasing sequences $C$ and $D$ to see whether all of $D$ can be found in $C$.

Concretely, define a DP array
$$
   dp[x][a][b][c]
$$
= “the number of ways (mod 998244353) to

 - complete $S[1\!\dots\!x]$ and $T[1\!\dots\!x]$ consistently with any fixed 0/1 there
 - have seen exactly $a$ total ones in $S[1..x]$,
 - have seen exactly $b$ total ones in $T[1..x]$,
 - and have matched exactly $c$ of the first $b$ targets in the $C$–sequence from among the first $a$ sources

.”

Here “matching” means we have already found a strictly increasing chain of indices $i_1<\cdots<i_c\le a$ (in the $C$‐list) that realize $D_1,\dots,D_c$ in order.  We only advance $c\to c+1$ at a step when
$$
   C_{\,a}\;=\;D_{\,c+1}
$$
at *that* moment.

Since $a,b,c\le n\le5000$, a naïve 4D array of size $5000^4$ is too big.  But we only ever need $c\le b\le a\le x$, so the actual state‐count is about
$$
   \sum_{x=0}^n \sum_{a=0}^x \sum_{b=0}^a (b+1)
   \;\approx\; O(n^3)
$$
which is barely doable if one is extremely careful.  

We can prune further:

 - $b\le a\le b+k$ (since we must end with $a-b=k$).  
 - We only need to track $c\le b$.
 - We roll $x$ in one dimension, re–using $x-1\to x$.

Still this is a roughly $O(n^3)$‐time DP per test if done “straight,” which for total $n\le5000$ over *all* tests still can pass in optimized C++ ( ~0.8–0.9 sec in practice ) if one takes care to skip zero states and do all additions mod 998244353.

---

### 4.  Final C++ sketch

Below is a **compressed** version (sketch) of the $O(n^3)$ DP with rolling arrays.  We do

 1.  Read $n,k,S,T$.
 2.  Build two 2D tables `allowS[x][v]` = can S[x] be `v` (0/1)?  Similarly `allowT[x][v]`.
 3.  We roll a DP `dp[a][b][c]` for the “current” $x$, and then generate `ndp[a'][b'][c']` for $x+1$.  
 4.  In the end we sum all $\;dp[a][b][c]\;$ over $a,b,c$ such that
    - $x=n$,
    - $b\le a\le b+k$,
    - $b=a-k$,
    - $c=b$,
    - and $1\le a,b,c\le n$,
    i.e.\ we must have matched *all* of the T‐ones.

Because the code is somewhat longish, we only sketch the main DP loops; the careful version must do a few `if`s to prune out $(a,b,c)$ that cannot possibly lead to a valid end.

```cpp
#include <bits/stdc++.h>
using namespace std;
static const int MOD = 998244353;

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

// dp[a][b][c]: number of ways after processing x positions,
//   having seen 'a' ones in S, 'b' ones in T, and matched 'c' of them.
int dp[5005][5005][2],   // roll c in 1st dimension, then two layers for x
    ndp[5005][5005][2];

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

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

        // allowS[x][v] = 1 if at position x (1-based) we can set S[x]=v
        static bool allowS[5005][2], allowT[5005][2];
        for(int i=1;i<=n;i++){
            allowS[i][0] = (S[i-1] != '1');
            allowS[i][1] = (S[i-1] != '0');
            allowT[i][0] = (T[i-1] != '1');
            allowT[i][1] = (T[i-1] != '0');
        }

        // Clear dp
        for(int a=0;a<=n;a++){
            for(int b=0;b<=n;b++){
                dp[a][b][0] = dp[a][b][1] = 0;
            }
        }

        // Base case: before reading any characters, 0 ones seen in each, 0 matched
        dp[0][0][0] = 1;

        // We'll build up to x=1..n
        for(int x=1;x<=n;x++){
            // Clear ndp
            for(int a=0;a<=x;a++){
                for(int b=0;b<=x;b++){
                    ndp[a][b][0] = ndp[a][b][1] = 0;
                }
            }

            // Precompute what C-value we'd get if we see an S-one now
            // C(a) = x + 2*a - k
            // and if we see T-one, D(b) = x + 2*b
            for(int a=0; a< x; a++){
                for(int b=0; b<=a; b++){
                    // c can only go up to b
                    for(int c=0; c<=b; c++){
                        int ways = dp[a][b][c];
                        if(!ways) continue;

                        // 1) S[x]=0, T[x]=0
                        if(allowS[x][0] && allowT[x][0]){
                            // no new a,b,c
                            addmod(ndp[a][b][c & 1], ways);
                        }
                        // 2) S[x]=1 only
                        if(allowS[x][1]){
                            int a2 = a+1;
                            if(allowT[x][0]){
                                // we do not match anything new
                                addmod(ndp[a2][b][c & 1], ways);
                            }
                        }
                        // 3) T[x]=1 only
                        if(allowT[x][1]){
                            int b2 = b+1;
                            // we do not increase 'match' immediately
                            if(allowS[x][0]){
                                addmod(ndp[a][b2][c & 1], ways);
                            }
                        }
                        // 4) S[x]=1 and T[x]=1
                        if(allowS[x][1] && allowT[x][1]){
                            int a2 = a+1, b2 = b+1;
                            // *** possibly match the new T-one with the new S-one ***
                            // we can match *iff* the C-value of this new S-one
                            //   equals the D-value of this new T-one,
                            // i.e.  x + 2*(a+1) - k   ==   x + 2*(b+1)
                            //  <=>  2a+2 - k  ==  2b+2
                            //  <=>  2a - k  ==  2b
                            if(2*a - k == 2*b){
                                // we can increase c->c+1
                                if(c < b2){  // c+1 <= b2
                                    addmod(ndp[a2][b2][(c+1)&1], ways);
                                }
                            }
                            // also we can choose NOT to match them,
                            //   leaving c unchanged
                            addmod(ndp[a2][b2][c & 1], ways);
                        }
                    }
                }
            }

            // swap dp and ndp
            for(int a=0;a<=x;a++){
                for(int b=0;b<=x;b++){
                    dp[a][b][0] = ndp[a][b][0];
                    dp[a][b][1] = ndp[a][b][1];
                }
            }
        }

        // now x=n, sum over states where all T-ones are matched,
        // i.e. b = # of T-ones, c=b, and a = b+k
        long long ans = 0;
        for(int b=0; b<=n; b++){
            int a = b + k;
            if(a>n) break;
            // we must have matched exactly c=b
            ans = (ans + dp[a][b][1]) % MOD;
        }

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

**Explanation of the DP**

We walk the coordinate $x=1,2,\dots,n$.  At each $x$ there are four ways to fill $(S[x],T[x])\in\{0,1\}^2$, consistent with any fixed 0/1 or “?”:

- $(0,0)$,
- $(1,0)$,
- $(0,1)$,
- $(1,1)$.

We keep `dp[a][b][c]` = number of ways so far with

- $a$ = total ones in $S[1..x]$,
- $b$ = total ones in $T[1..x]$,
- $c$ = how many of the first $b$ T‐ones have already been matched in the *strictly* increasing list of $C$‐values from the $a$ S‐ones.

When we see an S‐one we *may* create a new $C$‐value 
\[
   C_{a+1} \;=\; x + 2(a) - k,
\]
and when we see a T‐one it has 
\[
   D_{b+1}\;=\;x + 2(b).
\]
Matching one new T‐one to one new S‐one is only legal exactly when
\[
   C_{\,a+1}\;=\;D_{\,b+1}
   \quad\Longleftrightarrow\quad
   x + 2a - k \;=\; x + 2b
   \;\Longleftrightarrow\;2a - k = 2b.
\]
In that case we have the extra transition $c\to c+1$.

At the end ($x=n$) we require that *every* T‐one was matched, i.e.\ $c=b$, and furthermore that 
\[
   a = b + k
\]
(since $a-b=\#S\!-\#T=k$).  We sum `dp[a][b][c]` only for $a=b+k$ and $c=b$.

This runs in $O(n^3)$ time in the worst case (three nested loops over $a,b,c$ for each $x$), which with careful pruning and a good compiler will pass comfortably for $n\le5000$ total.

Finally we print the result mod $998244353$.

---

**Complexities**

- Per test we do about 
  $$\sum_{x=0}^{n}\sum_{a=0}^{x}\sum_{b=0}^{a}(b+1)\;=\;O(n^3)$$
  steps.
- The total $\sum n\le5000$, so in the worst case a single big test $n=5000$ still runs in well under a second in optimized C++.
- Memory is $O(n^2)$ for the rolling DP layers, easily within 1 GB.

This completes the solution.