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 N=2e5+5,YL=1e9+7;
inline int MO(const int &x){return x>=YL?x-YL:x;}
inline int in(){int k;cin>>k;return k;}
inline int low(int x){return x&(-x);}
char s[N];
typedef std::pair<int,int> P;
#define mk make_pair
int f[205][N],a[N];
inline void add(int &x,int y){if((x+=y)>=YL)x-=YL;}
int main()
{
	ios::sync_with_stdio(0);
	int n,m,ans=0;cin>>n>>m;
	for(int i=1;i<=n;++i)a[i]=in();
	std::sort(a+1,a+n+1);f[0][m]=1;
	std::reverse(a+1,a+n+1);
	for(int i=1;i<=n;++i)
		for(int j=0;j<N;++j)
			add(f[i][j%a[i]],1ll*f[i-1][j]),add(f[i][j],1ll*f[i-1][j]*(n-i)%YL);
	for(int i=0;i<N;++i)add(ans,1ll*i*f[n][i]%YL);
	printf("%d\n",ans);
	return 0;
}
