Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

2 19
3 7

Sample Output 1

3

There are two possible orders in which we remove the numbers from S.

If we remove 3 and 7 in this order, the number on the blackboard changes as follows: 19 \rightarrow 1 \rightarrow 1.

If we remove 7 and 3 in this order, the number on the blackboard changes as follows: 19 \rightarrow 5 \rightarrow 2.

The output should be the sum of these: 3.

Sample Input 2

5 82
22 11 6 5 13

Sample Output 2

288

Sample Input 3

10 100000
50000 50001 50002 50003 50004 50005 50006 50007 50008 50009

Sample Output 3

279669259

Be sure to compute the sum modulo 10^{9}+7.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define forx(i, a, b) for (int i = (a); i < (b); i++)
#define rep(i, n) for (int i = 0; i < (n); i++)

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n,x,mod=1e9+7;
    cin>>n>>x;
    vector<int>s(n);
    rep(i,n)cin>>s[i];

    ll dp[n+1][x+1]={0};
    rep(i,x)dp[0][i]=x;

    rep(i,n)rep(j,x+1){
        dp[i+1][j]+=dp[i][x]%mod;
        dp[i+1][x]%=mod;
        dp[i+1][j]+=dp[i][x%s[i]]%mod;
        dp[i+1][x]%=mod;
    }

    cout<<dp[n][x]<<endl;

    return 0;
}