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 <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
ll N, X;
ll S[200];
ll d[201][100001];
ll Ns;

ll stair(ll x){
    if(x==1){
        return 1;
    }
    return x*stair(x-1);
}

int main(void) {
    // insert code here...
    cin >> N >> X;
    
    Ns = stair(N);
//    cout << Ns << endl;
    
    for(ll i = 0; i < N; i++){
        cin >> S[i];
    }
    
    sort(S, S+N);
    reverse(S, S+N);
    
    for(ll i = 0; i < X+1; i++){
        d[N][i] = i*Ns;
//        cout << d[N][i] << " ";
    }
//    cout << endl;
    
    for(ll i = N-1; i >= 0; i--){
        for(ll j = 0; j < X+1; j++){
            
            d[i][j] = d[i+1][j%S[i]]/(N-i) + d[i+1][j]*(N-i-1)/(N-i);
            
//            if(d[i][j] > 1000000007){
//                d[i][j] = d[i][j]%1000000007;
//            }
//            cout << d[i][j] << " ";
        }
//        cout << endl;
    }
    
    cout << d[0][X]%1000000007 << endl;
    
    return 0;
}