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<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int mod=1000000007;
const int N=100050;
int a[205],f[205],d[205],dp[205][N];
void exgcd(int a,int b,int &x,int &y){
	if(!b){x=1;y=0;}
	else{
		exgcd(b,a%b,y,x);
		y-=a/b*x;
	}
}
inline int inv(int a){
	int x,y;
	exgcd(a,mod,x,y);
	return x<0?x+mod:x;
}
int dfs(int m,int x){
	if(!m)return x;
	if(x<a[1])return (ll)x*f[m]%mod;
	if(dp[m][x]!=-1)return dp[m][x];
	int i,ans=0;
	for(i=1;i<=m;++i)ans=(ans+(ll)dfs(i-1,x%a[i])*f[m-1]%mod*d[i-1])%mod;
	return dp[m][x]=ans;
}
int main(){
	int n,x,i;
	scanf("%d%d",&n,&x);
	for(i=1;i<=n;++i)scanf("%d",a+i);
	sort(a+1,a+n+1);
	f[0]=f[1]=d[0]=d[1]=1;
	for(i=2;i<n;++i){
		f[i]=(ll)f[i-1]*i%mod;
		d[i]=inv(f[i]);
	}
	memset(dp,-1,sizeof(dp));
	printf("%d\n",dfs(n,x));
	return 0;
}