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>
using namespace std;
typedef long long ll;
typedef long double ld;

const int MAXS = 1e5+10;
const int MAXN = 222;
const ll M = 1e9+7;
int n, x;

ll mem[MAXN][MAXS], s[MAXN];

ll dp(int tot, int x) { // total number chosen nodes , last x
  ll & ans = mem[tot][x];
  if (ans!=-1) return ans;
  if (tot==n) return x;
  ans = 0;
  int i;
  for(i=0;s[i]<=x && i<n;i++) {
    ans = (ans+dp(tot+1, x%s[i]))%M;
  }
  if (n-i>tot) {
    ans = (ans+(n-i-tot)*dp(tot+1,x))%M;
  }
  return ans;
}

int main(){
  cin.tie(0); ios_base::sync_with_stdio(0);
  cin >> n >> x;
  memset(mem,-1, sizeof mem);
  for(int i=0;i<n;i++) cin >> s[i];
  sort(s, s+n);
  cout << dp(0, x) <<endl;
}
