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 fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

typedef long long LL;

const int maxn=205, maxx=1e5+5;
const LL mo=1e9+7;

int n,x,a[maxn];

bool cmpA(int a,int b) {return a>b;}

LL f[maxn][maxx];
int main()
{
	scanf("%d %d",&n,&x);
	fo(i,1,n) scanf("%d",&a[i]);
	sort(a+1,a+1+n,cmpA);
	
	f[0][x]=1;
	fo(i,0,n-1)
	{
		fo(j,0,x)
		{
			(f[i+1][j]+=f[i][j]*(n-i-1))%=mo;
			(f[i+1][j%a[i+1]]+=f[i][j])%=mo;
		}
	}
	
	LL ans=0;
	fo(j,0,x) (ans+=f[n][j]*j)%=mo;
	
	printf("%lld\n",ans);
}