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 mod = 1000000007;
int a[2000], fct[2000];
int f[201][100001];
int main() {
	ios_base::sync_with_stdio(false);
	int n, x;
	cin >> n >> x;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
	}
	sort(a + 1, a + n + 1);
	f[n][x] = 1;
	for (int i = n; i >= 1; i--) {
		for (int j = x; j >= 0; j--) {
			if (f[i][j] == 0) {
				continue;
			}
			f[i - 1][j % a[i]] += f[i][j];
			if (f[i - 1][j % a[i]] >= mod) {
				f[i - 1][j % a[i]] -= mod;
			}
			f[i - 1][j] += 1ll * f[i][j] * (i - 1) % mod;
			
			if (f[i - 1][j] >= mod) {
				f[i - 1][j] -= mod;
			}
		}
	}
	int ans = 0;
	for (int i = 0; i < a[1]; i++) {
		ans += 1ll * i * f[0][i] % mod;
		if (ans >= mod) {
			ans -= mod;
		}
	}
	cout << ans << "\n";
	return 0;
}