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 <bits/stdc++.h>
#define st first
#define nd second
#define mp make_pair
#define pb push_back
#define mod 1000000007
#define N 100005
using namespace std;

typedef long long ll;

ll n, x, ans, a[N], dp[205][N];

int main() {
	// freopen("in.txt", "r", stdin);
	// freopen("out.txt", "w", stdout);

	scanf("%lld %lld",&n ,&x);

	for(int i = 1; i <= n; i++)
		scanf("%lld", a + i);

	sort(a + 1, a + n + 1);
	reverse(a + 1, a + n + 1);
	a[n + 1] = 1;
	dp[1][x] = 1;
	for(int i = 1; i <= n + 1; i++)
		for(int j = 0; j < N; j++){
			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]*(n - i)%mod )%mod;
			// printf("dp[%lld][%lld] = %lld\n", i, j, dp[i][j]);
		}
	for(int j = 0; j < N; j++)
		ans = (ans + j*dp[n + 1][j]%mod)%mod;
	printf("%lld\n", ans);
	return 0;
}