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 mod=1e9+7;
const int N=1e5+10;
int n,X;
int a[N];
int dp[2][N];
ll ans;
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);
	int now=0,lst=1;
	dp[lst][X]=1;
	for(int i=1;i<=n;i++,now^=1,lst^=1){
		memset(dp[now],0,sizeof(dp[now]));
		if(i!=n)for(int j=0;j<=X;j++)dp[now][j]=(dp[now][j]+(ll)(n-i)*dp[lst][j]%mod)%mod;
		for(int j=0;j<=X;j++)dp[now][j%a[i]]=(dp[now][j%a[i]]+dp[lst][j])%mod;
	}
	for(int i=1;i<=X;i++)ans=(ans+(ll)i*dp[lst][i]%mod)%mod;
	ans=(ans%mod+mod)%mod;
	printf("%lld\n",ans);
	return 0;
}