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 MOD 1000000007LL
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int n,x;
ll dp[205][200005];
int s[205];

ll extgcd(ll a,ll b,ll& x,ll& y){
    ll d=a;
    if(b!=0LL){
        d=extgcd(b,a%b,y,x);
        y-=(a/b)*x;
    }else{
        x=1;
        y=0;
    }
    return d;
}

ll mod_inverse(ll a,ll m){
    ll x,y;
    extgcd(a,m,x,y);
    return (m+x%m)%m;
}

int main(void){
    scanf("%d%d",&n,&x);
    for(int i=0;i<n;i++){
        scanf("%d",&s[i]);
    }
    sort(s,s+n);
    for(int i=0;i<=x;i++){
        dp[0][i]=i;
    }
    for(int i=0;i<n;i++){
        for(int j=0;j<=x;j++){
            dp[i+1][j]=(ll)dp[i][j]*i%MOD;
            dp[i+1][j]+=dp[i][j%s[i]];
            dp[i+1][j]%=MOD;
        }
    }
    printf("%lld\n",dp[n][x]);
    return 0;
}