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 N 200005

using namespace std;

const int Mod=1e9+7; 
int n,w,a[N],dp[205][N];

int main(){
    scanf("%d%d",&n,&w);
    for (int i=1;i<=n;i++) scanf("%d",&a[i]);
    sort(a+1,a+n+1);reverse(a+1,a+n+1);
    dp[0][w]=1;
    for (int i=0;i<n;i++){
        for (int j=0;j<=w;j++){
            dp[i+1][j%a[i+1]]=(dp[i+1][j%a[i+1]]+dp[i][j])%Mod;
            dp[i+1][j]=(dp[i+1][j]+1ll*dp[i][j]*(n-i-1)%Mod)%Mod;
        }
    }
    int ans=0;
    for (int i=0;i<a[n]&&i<=w;i++) ans=(ans+1ll*dp[n][i]*i%Mod)%Mod;
    printf("%d\n",ans);
    return 0;
}
//uoj