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;
#define REP(i,N) for(i=0;i<N;i++)
typedef long long ll;
typedef pair<int, int> P;
typedef struct{
  int first;
  int second;
  int third;
}T;

//昇順
bool comp_Se(T& l, T& r){
  return l.second < r.second;
}

ll mod =  1e9+7;
int main(void){
  int i,j,N,X;
  cin >> N >> X;
  vector<ll> v(N);
  REP(i,N) cin >> v[i];
  sort(v.begin(),v.end());

  //dp[i][j]:jをXの最初として，i+1回操作(ただしi+1番目まで)を行った時の総和
  ll dp[N][X+1];
  REP(j,X+1){
    dp[0][j] = j%v[0];
  }
  for(i=1;i<N;i++){
    REP(j,X+1){
      //dp[i-1][j%v[i]]:最初に行う場合
      //dp[i-1][j]*i 最初に行わない場合
      dp[i][j] = dp[i-1][j%v[i]] + dp[i-1][j] * i; 
      dp[i][j] %= mod;
    }
  }
  cout << dp[N-1][X] << endl;
  return 0;
}
