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 <algorithm>

using namespace std;

typedef long long ll;

ll n, w, dp[205][100005], a[205];
const ll mod = (ll)1e9 + 7;

bool cmp(ll x, ll y){
	return x > y;
}

int main(){
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n >> w;
for(int i = 1 ; i <= n ; ++i){
	cin >> a[i];
}
sort(a + 1, a + n + 1, cmp);
dp[0][w] = 1;
for(int i = 1 ; i <= n ; ++i){
	for(int j = 0 ; j <= w ; ++j){
		dp[i][j % a[i]] = (dp[i][j % a[i]] + dp[i - 1][j]) % mod;
		dp[i][j] = (dp[i][j] + (dp[i - 1][j] * (n - i)) % mod) % mod;
	}
}
ll sol = 0;
for(int i = 0 ; i <= w ; ++i){
	sol = (sol + (dp[n][i] * i) % mod) % mod;
}
cout << sol << endl;

return 0;
}

