#pragma comment(linker, "/stack:16777216")
#include <string>
#include <vector>
#include <map>
#include <list>
#include <iterator>
#include <set>
#include <queue>
#include <iostream>
#include <sstream>
#include <stack>
#include <deque>
#include <cmath>
#include <memory.h>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <algorithm>
#include <utility>

using namespace std;

#define FOR(i,a,b) for(int i = (a); i < (b); ++i)
#define RFOR(i,b,a) for(int i = (b) - 1; i >= (a); --i)
#define REP(i,N) FOR(i, 0, N)
#define RREP(i,N) RFOR(i, N, 0)
#define FILL(A,value) memset(A,value,sizeof(A))

#define ALL(V) V.begin(), V.end()
#define SZ(V) (int)V.size()
#define PB push_back
#define MP make_pair
#define Pi 3.14159265358979

typedef long long Int;
//typedef unsigned long long UINT;
typedef vector <int> VI;
typedef pair <int, int> PII;

const int INF = 1000000000;
const int MAX = 64;
const int MAX2 = 1000000;
const int BASE = 1000000000;
const int CNT = 20;

#define MOD 1000000007

struct Test
{
    int n;
    string s;
    string t;
    Test(){}
    int R[MAX];
    int solve()
    {
        int m = SZ(s);
        int maskS = 0;
        FOR (i,0,m)
            if (s[i] == 'B')
                maskS += (1 << (m-i-1));
        int maskT = 0;
        FOR (i,0,m)
            if (t[i] == 'B')
                maskT += (1 << (m-i-1));
        // For each length len, let R[len] be the number of strings of A and B of length len that are good (simply brute force)
        R[0] = 1;
        FOR (len,1,n+1)
        {
            R[len] = 0;
            if (len < m)
            {
                R[len] = (1 << len);
                continue;
            }
            FOR (mask,0,(1 << len))
            {
                bool ok = 1;
                for (int j = 0; j+m-1 < len; ++j)
                {
                    int t = (mask >> j) % (1 << m);
                    if (t >= maskS && t <= maskT)
                    {
                        ok = 0;
                        break;
                    }
                }
                R[len] += ok;
            }
        }
        Int res = 0;
        // Now lets iterate where non-AB characters are. All position between them can be filled using the array R.
        FOR (mask,0,(1 << n))
        {
            Int p = 1;
            int last = -1, cnt = 0;
            FOR (i,0,n)
            {
                if ((mask & (1 << i)) != 0)
                {
                    p = p * R[i-last-1];
                    last = i;
                    ++ cnt;
                }
            }
            p = p * R[n-last-1];
            FOR (i,0,cnt)
                 p = (p * 24) % MOD;
            res += p;
            res %= MOD;
        }
        return res;
    }
    void writeTest()
    {
        cout << n << ' ' << SZ(s) << endl;
        cout << s << endl;
        cout << t << endl;
    }
    void writeRes()
    {
        cout << solve() << endl;
    }
    void readTest()
    {
        int m;
        cin >> n >> m;
        cin >> s >> t;
    }
};

int main()
{
    int cnt;
    cin >> cnt;
    FOR (i,0,cnt)
    {
        Test t;
        t.readTest();
        t.writeRes();
    }





	return 0;
}