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 LL mod=1000000007;
LL n,m,ans,a[205],f[205][100005];

void upd(LL &x,LL y){x=(x+y)%mod;}

int main(){
	scanf("%lld%lld",&n,&m);
	for (LL i=1;i<=n;++i) scanf("%lld",a+i);
	sort(a+1,a+n+1,greater<LL>());
	f[0][m]=1;
	for (LL i=1;i<=n;++i)
		for (LL j=0;j<=100000;++j) if (f[i-1][j]){
			upd(f[i][j%a[i]],f[i-1][j]);
			upd(f[i][j],f[i-1][j]*(n-i));
		}
	for (LL i=0;i<=100000;++i) upd(ans,f[n][i]*i);
	printf("%lld\n",ans);
	
	return 0;
}
			