Problem Description



Xuejiejie loves strings most. In order to win the favor of her, a young man has two strings X, Y

The question is that : Define the L as the length of the longest common subsequence of X and Y.( The subsequence does not need to be continuous in the string, and a string of length L has 2L subsequences containing the empty string ). Now Xuejiejie comes up with all subsequences of length L of string X, she wants to know the number of subsequences which is also the subsequence of string Y.


Input



In the first line there is an integer T, indicates the number of test cases.

In each case:

The first line contains string X, a non-empty string consists of lowercase English letters.

The second line contains string Y, a non-empty string consists of lowercase English letters.

1≤|X|,|Y|≤1000, |X| means the length of X.


Output



For each test case, output one integer which means the number of subsequences of length L of X which also is the subsequence of string Y modulo 109+7.


Sample Input 



2
a
b
aa
ab


Sample Output



1

2

dp

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<queue>
using namespace std;
const int maxn = 1005;
const __int64 base = 1000000007;
int n, m, T, na, nb;
char a[maxn], b[maxn];
__int64 f[maxn][maxn], sum[maxn][maxn], ans;

int main()
{
    scanf("%d", &T);
    while (T--)
    {
        scanf("%s%s", a, b);
        na = strlen(a);
        nb = strlen(b);
        memset(f, 0, sizeof(f));
        memset(sum, 0, sizeof(sum));
        for (int i = 0; i <= na; i++) sum[i][0] = 1;
        for (int i = 0; i <= nb; i++) sum[0][i] = 1;
        for (int i = 1; i <= na; i++)
            for (int j = 1; j <= nb; j++)
            {
                if (a[i - 1] == b[j - 1])
                {
                    f[i][j] = f[i - 1][j - 1] + 1;
                    sum[i][j] = sum[i - 1][j - 1];
                    if (f[i][j] == f[i - 1][j]) sum[i][j] += sum[i - 1][j];
                    //if (f[i][j] == f[i][j - 1]) sum[i][j] += sum[i][j - 1];
                }
                else
                {
                    f[i][j] = max(f[i - 1][j], f[i][j - 1]);
                    if (f[i][j] == f[i - 1][j]) sum[i][j] += sum[i - 1][j];
                    if (f[i][j] == f[i][j - 1]) sum[i][j] += sum[i][j - 1];
                    if (f[i][j - 1] == f[i - 1][j] && f[i][j] == f[i - 1][j - 1])
                        sum[i][j] -= sum[i - 1][j - 1];
                }
                sum[i][j] %= base;
            }
        //printf("%I64d\n", f[na][nb]);
        (sum[na][nb] += base) %= base;
        printf("%I64d\n", sum[na][nb]);
    }
    return 0;
}