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 <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;

ll mod=1e9+7;
ll fac[210];

void modfac(ll n){
    fac[0]=1;
    for(ll i=1;i<n;i++){
        fac[i]=(fac[i-1]*i)%mod;
    }
}

ll modinv(ll a){
    ll b=mod,u=1,v=0;
    while(b){
        ll t=a/b;
        a-=t*b; swap(a,b);
        u-=t*v; swap(u,v);
        }
        u%=mod;
        if(u<0) u+=mod;
    return u;
}

ll dp[210][100010];

int main(){
  ll n,x;
  cin >> n >> x;
  ll a[n];
  for(ll i=0;i<n;i++) cin >> a[i];
  sort(a,a+n);
  reverse(a,a+n);
  ll inv[n+1];
  for(ll i=1;i<=n;i++) inv[i]=modinv(i);
  for(ll i=0;i<=x;i++) dp[n][i]=i;
  for(ll i=n-1;i>=0;i--){
    for(ll j=0;j<=x;j++){
      ll p=n-i;
      dp[i][j]=inv[p]*(dp[i+1][j%a[i]]+(p-1)*dp[i+1][j]%mod)%mod;
    }
  }
  modfac(210);
  cout << dp[0][x]*fac[n]%mod << endl;
}