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 <iostream>
#include <vector>
#include <algorithm>

using namespace std;
typedef long long ll;
const int MAX = 100010;
ll inv[MAX],fact[MAX],fi[MAX],mod = 1000000007;
vector<int> s;
void inverse(){
	int i;
	inv[1] = 1;
	for(i=2;i<MAX;i++){
		inv[i] = mod - (mod/i)*inv[mod%i]%mod;
	}
	fact[0] = fi[0] = 1;
	for(i=1;i<MAX;i++){
		fact[i] = fact[i-1]*i%mod;
		fi[i] = (fi[i-1]*inv[i])%mod;
	}
}

ll ex[210][100010] = {};
int main(){
	int i,j,n,x;
	cin >> n >> x;
	for(i=0;i<n;i++){
		int y; cin >> y;
		s.push_back(y);
	}
	inverse();
	sort(s.rbegin(),s.rend());
	for(i=0;i<=x;i++){
		ex[n][i] = i;
	}
	for(i=n - 1;i>=0;i--){
		for(j=0;j<=x;j++){
			(ex[i][j] += (inv[n - i]*ex[i + 1][j%s[i]]%mod + (mod + 1 - inv[n - i])*ex[i + 1][j]%mod)) %= mod;
		}
	}
	cout << ex[0][x]*fact[n]%mod << endl;
}