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: Accepted)

#include<bits/stdc++.h> 
using namespace std;
using ll=long long;
typedef pair<ll,ll> P;
#define all(v) (v).begin(),v.end()
#define fi first
#define se second
const ll inf=(1e18);
const ll mod=1000000007;
ll dp[300][200000];
int main(){
int n,x;
cin>>n>>x;
vector<ll> a(n);
for(ll i=0;i<n;i++)cin>>a[i];
dp[0][x]=1;
sort(all(a),greater<ll>());
for(ll i=0;i<n;i++){
    for(ll j=0;j<=x;j++){
            dp[i+1][j%a[i]]+=dp[i][j];
            dp[i+1][j%a[i]]%=mod;          
            dp[i+1][j]+=dp[i][j]*(n - 1 - i);
            dp[i+1][j]%=mod;
         }
 }
 ll ans=0;
 for(ll i=1;i<=x;i++){
     ans+=dp[n][i]*i;
     ans%=mod;
}
cout<<ans<<endl;
}