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>
using namespace std;
typedef long long LL;
const int N=200+10;
const int M=1e5+10;
const int MOD=1e9+7;
int n,x;
int a[N],f[N][M];

int main()
{
	scanf("%d%d",&n,&x);
	for (int i=1;i<=n;i++) scanf("%d",&a[i]);
	sort(a+1,a+n+1);
	reverse(a+1,a+n+1);
	
	f[1][x]=1;
	for (int i=1;i<=n;i++)
		for (int j=0;j<=x;j++) {
			f[i+1][j%a[i]]=((LL)f[i+1][j%a[i]]+(LL)f[i][j])%MOD;
			f[i+1][j]=((LL)f[i+1][j]+(LL)f[i][j]*(n-i)%MOD)%MOD;
		}
	
	LL ans=0;
	for (int j=0;j<x;j++) ans=(ans+(LL)f[n+1][j]*j)%MOD;
	cout<<ans<<endl;	
	return 0;
}