Shared code

#include<bits/stdc++.h>
#define all(x) x.begin(), x.end()
#define ll long long
using namespace std;
template <typename T>
using bstring = basic_string<T>;
template <typename T>
using matrix = vector<vector<T>>;

const int MOD = 1e9 + 7;

int main(){
    
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int n, m;
    cin >> n >> m;
 
    matrix <int> comp (1 << n, vector <int> (1<<n));
 
 
    // checks for every pair of masks if they are compatible
    for(int i = 0; i < (1<<n); i++){
        for(int j = 0; j < (1<<n); j++){
            int innactive = (~i)&(~j); // here a bit is on iff its inactive in both i and j
            int last = -1; //last = least significant bit of the current segment of bits that are innactive in both i and j
            // last = -1 is a code for not being on a segment of that sort 
            bool ok = 1;
            for(int k = 0; k < n; k++){
                if((1<<k)&innactive){
                    if(last == -1) // we enter a segment
                        last = k;
                } else {
                    if(last != -1){ // we leave a segment
                        ok&=(k-last)%2 == 0;
                        last = -1;
                    }
                }
            }
            if(last!=-1)
                ok&=(n-last)%2 == 0;
            
            comp[i][j] = ok;
        }
    }
 
 
    matrix<ll> dp(m+2,vector<ll>(1<<n));
 
    dp[1][0] = 1; // innitially theres nothing poking into column 1
 
    for(int i = 2; i <= m+1; i++){
        for(int mask = 0; mask < (1<<n); mask++){
            int poss = (~mask)&((1<<n)-1); // sets up ~X' so that we can search X that are submasks of it
 
            for(int sub = poss; sub; sub = (sub-1)&poss){
                //note that comp[a][b] == comp[b][a]
                dp[i][mask] += dp[i-1][sub]*comp[mask][sub];
            }
            dp[i][mask]+=dp[i-1][0]*comp[mask][0]; //our previous for loop doesnt go over mask 0, so we have to check it mannually
            dp[i][mask]%=MOD;
        }
    }
 
    cout << dp[m+1][0] << '\n';
 
    
    return 0;
 
}