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 <cstdlib>
#include <algorithm>
#include <vector>
#include <cstring>
#include <string>
#include <queue>
#include <functional>
#include <set>
#include <map>
#include <deque>
#include <cmath>
#include <cassert>
#define SIZE 205
#define MX 100005
#define MOD 1000000007

using namespace std;
typedef long long int ll;
typedef pair <int,int> P;

int A[SIZE];
ll dp[2][MX];

int main()
{
	int n,X;
	scanf("%d %d",&n,&X);
	for(int i=0;i<n;i++) scanf("%d",&A[i]);
	sort(A,A+n);
	int pos=0;
	for(int i=0;i<MX;i++) dp[pos][i]=i;
	for(int i=0;i<n;i++)
	{
		pos^=1;
		for(int j=0;j<MX;j++)
		{
			dp[pos][j]=dp[pos^1][j]*(ll) i%MOD;
			dp[pos][j]+=dp[pos^1][j%A[i]];
			if(dp[pos][j]>=MOD) dp[pos][j]-=MOD;
		}
	}
	printf("%lld\n",dp[pos][X]);
	return 0;
}
