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
// I SELL YOU...! 
#include<iostream>
#include<vector>
#include<algorithm>
#include<functional>
#include<queue>
#include<chrono>
#include<iomanip>
#include<map>
#include<set>
using namespace std;
using ll = long long;
using P = pair<ll,ll>;
void init_io(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << setprecision(10);
}
#define MOD 1000000007
#define MAX 110000
ll dp[220][MAX]={};
signed main(){
  init_io();
  ll n,x,ans=0;
  cin >> n >> x;
  vector<ll> s(n);
  for(int i=0;i<n;i++){
    cin >> s[i];
  }
  sort(s.begin(),s.end());
  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] += dp[i][j%s[i]];
      dp[i+1][j] %= MOD;
      dp[i+1][j] += dp[i][j] * i;
      dp[i+1][j] %= MOD;
    }
  }
  cout << dp[n][x]<<endl;
}
