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 <iostream>
#include <vector>
#include <algorithm>
#include <utility>
#include <queue>
#include <string>
#include <set>
#include <bitset>
#include <unordered_map>

using namespace std;
using ll = long long;
using Pii = pair<int, int>;
using Pll = pair<ll, ll>;
template <class T>
using Vvec = vector<vector<T> >;

int main()
{
    const ll MOD = 1e9+7; 
    int N, X;
    cin >> N >> X;
    vector<ll> S(N);
    for(ll& x: S) cin >> x;
    sort(S.begin(), S.end());

    vector<ll> dp(X+1);
    for(int j=0; j<=X; j++) dp[j] = j % S[0];
    for(int i=1; i<N; i++){
        for(int j=X; j>=0; j--){
            dp[j] = dp[j%S[i]] + dp[j] * i;
            dp[j] %= MOD;
        }
    }
    cout << dp[X] << endl;
}