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>
#define rep(i,n) for(int i=(0);i<(n);i++)

using namespace std;

typedef long long ll;

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

	ll MOD = 1e9+7;

	int N, X;
	cin >> N >> X;

	vector<ll> S(N, 0);
	rep(i, N) cin >> S[i];

	sort(S.begin(), S.end(), greater<ll>());

	vector<vector<ll>> dp(N+1, vector<ll>(X+1, 0));
	dp[0][X] = 1;

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

	ll ans = 0;
	for(int i = 0; i <= X; i++){
		ans += i * dp[N][i] % MOD;
		ans %= MOD;
	}

	cout << ans << endl;
}