Snuke has a blackboard and a set S consisting of N integers. The i-th element in S is S_i.

He wrote an integer X on the blackboard, then performed the following operation N times:

* Choose one element from S and remove it.
* Let x be the number written on the blackboard now, and y be the integer removed from S. Replace the number on the blackboard with x \bmod {y}.



There are N! possible orders in which the elements are removed from S. For each of them, find the number that would be written on the blackboard after the N operations, and compute the sum of all those N! numbers modulo 10^{9}+7.

Constraints

* All values in input are integers.
* 1 \leq N \leq 200
* 1 \leq S_i, X \leq 10^{5}
* S_i are pairwise distinct.

Input

Input is given from Standard Input in the following format:


N X
S_1 S_2 \ldots S_{N}


Output

Print the answer.

Examples

Input

2 19
3 7


Output

3


Input

5 82
22 11 6 5 13


Output

288


Input

10 100000
50000 50001 50002 50003 50004 50005 50006 50007 50008 50009


Output

279669259
#include <bits/stdc++.h>
#define For(i, a, b) for(int (i)=(a); (i)<(b); ++(i))
#define rFor(i, a, b) for(int (i)=(a)-1; (i)>=(b); --(i))
#define rep(i, n) For((i), 0, (n))
#define rrep(i, n) rFor((i), (n), 0)
#define fi first
#define se second
using namespace std;
typedef long long lint;
typedef pair<int, int> pii;
const lint mod = 1e9 + 7;

int s[210];
lint dp[210][100010];

lint solve(int i, int j){
    if(dp[i][j] >= 0) return dp[i][j];
    return dp[i][j] = ((lint)(i-1)*solve(i-1, j) + solve(i-1, j%s[i-1])) % mod;
}

int main(){
    int n, x;
    scanf("%d%d", &n, &x);
    rep(i, n) scanf("%d", &s[i]);
    sort(s, s+n);
    rep(i, n+1)rep(j, x+1) dp[i][j] = -1;
    rep(j, x+1) dp[0][j] = j;
    printf("%lld\n", solve(n, x));
}