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<cstdio>
#include<algorithm>
int const x_max=100001;
//dp[x]=大きいほうから順に走査していって、xが出てくる場合の数
long long dp[x_max],dp_old[x_max];
long long const mod = 1e9+7;
int main(){
    int n,x,s[200];
    scanf("%d%d",&n,&x);
    for(int i=0;i<n;i++)scanf("%d",&s[i]);
    std::sort(s,s+n);
    for(int i=0;i<x;i++)dp[0]=0;
    dp[x]=1;
    //小さいほうからi番目を使うかどうかを、大きいほうから決める
    for(int i=n-1;i>=0;i--){
        //初期化
        for(int j=0;j<=x;j++){
            dp_old[j]=dp[j];
            dp[j]=0;
        }
        for(int j=0;j<=x;j++){
            //i番目を使う場合
            dp[j%s[i]]+=dp_old[j];
            //i番目を使わない場合
            dp[j]+=dp_old[j]*i;
        }
        //後処理
        for(int j=0;j<=x;j++)dp[j]%=mod;
    }
    long long ans=0;
    for(int i=0;i<=x;i++)ans+=i*dp[i];
    printf("%lld\n",ans%mod);
    return 0;
}