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.

Examples

Input

2 19
3 7


Output

3


Input

5 82
22 11 6 5 13


Output

288


Input

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


Output

279669259
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;
const int MOD = 1e9 + 7;
const int N = 205;
const int M = 1e5 + 10;
long long dp[N][M];
int a[N];

int main() {
	int n, m;
	cin >> n >> m;
	for(int i = 1; i <= n; i ++) {
		scanf("%d", &a[i]);
	}
	
	sort(a + 1 , a + 1 + n);

	dp[n][m] = 1;
	for(int i = n; i >= 1; i --) {
		for(int j = M - 5; j >= 0; j --) {
			if(dp[i][j] == 0) continue;
		   	dp[i - 1][j % a[i]] = (dp[i - 1][j % a[i]] +  dp[i][j]) % MOD;
			dp[i - 1][j] = (dp[i - 1][j] + dp[i][j] * 1LL * (i - 1)) % MOD;	
		}
	}

	long long sum = 0;
	for(int i = M - 5; i >= 0; i --) {
		sum = (sum + dp[0][i] * 1LL * i) % MOD;
	}
	cout << sum << endl;
	
}	
