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;
const int mod=1e9+7;
inline void add(int &a,int b){if((a+=b)>=mod)a-=mod;}
int n,p,a[601],f[220001][601],lim[220001];
int main()
{
	cin>>n>>p;
	for(int i=1;i<=n;i++)cin>>a[i];
	sort(a+1,a+1+n,greater<int>());
	for(int i=0;i<=p;i++)lim[i]=lower_bound(a+1,a+1+n,i,greater<int>())-a-1;
	f[p][lim[p]]=1;
	for(int x=p;x>=0;x--)
		for(int i=n;i>=0;i--)
			if(f[x][i])
			{
				if(i)add(f[x][i-1],1ll*f[x][i]*i%mod);
				for(int j=lim[x]+1;j<=n;j++)add(f[x%a[j]][lim[x%a[j]]-(lim[x]-i)-1],f[x][i]);
			}
	int ans=0;
	for(int x=0;x<a[n];x++)add(ans,1ll*x*f[x][0]%mod);
	cout<<ans<<'\n';
}
