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;
#define rep(i,n) for(int i = 0;i<((int)(n));i++)
#define reg(i,a,b) for(int i = ((int)(a));i<=((int)(b));i++)
#define irep(i,n) for(int i = ((int)(n)-1);i>=0;i--)
#define ireg(i,a,b) for(int i = ((int)(b));i>=((int)(a));i--)
typedef long long ll;
typedef pair<ll, ll> mp;


/*
AC
最大要素が先頭に来るかどうかで場合分け
*/

ll n,x,s[210],dp[210][100010]={},mod=1e9+7;

ll f(ll i,ll x){
	if(i==n)return x;
	if(dp[i][x]!=-1)return dp[i][x];
	ll ans=0;
	ans+=f(i+1,x%s[i]);
	ans+=f(i+1,x) * (n-1-i); 
	return dp[i][x]=ans%mod;
}

int main(void){
	cin>>n>>x;
	rep(i,n)cin>>s[i];
	sort(s,s+n);
	reverse(s,s+n);
	rep(i,210)rep(j,100010)dp[i][j]=-1;
	cout<<f(0,x)<<endl;
	return 0;
}