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 <algorithm>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <queue>
#include <iostream>
#include <bitset>
using namespace std;
#define N 100005
#define ll long long
#define mod 1000000007
int f[205][N],n,a[205],X,ans;
int main()
{
	scanf("%d%d",&n,&X);f[0][X]=1;
	for(int i=1;i<=n;i++)scanf("%d",&a[i]);sort(a+1,a+n+1);
	for(int i=1;i<=n;i++)
		for(int j=0,now=0;j<=X;j++)if(f[i-1][j])
		{
			while(now<n&&a[now+1]<=j)now++;
			if(n-now-i+1>0)f[i][j]=(f[i][j]+(ll)(n-now-i+1)*f[i-1][j])%mod;
			for(int k=1;k<=now;k++)(f[i][j%a[k]]+=f[i-1][j])%=mod;
		}
	for(int i=0;i<=X;i++){ans=(ans+(ll)f[n][i]*i)%mod;}
	printf("%d\n",ans);
}
