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 <map>
#include <set>
#include <queue>
#include <string>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <stdio.h>
using namespace std;
#define int long long
int MOD = 1000000007;
int dp[100001];
int ndp[100001];
signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N, X;
	cin >> N >> X;
	vector<int> A(N);
	int res = 0;
	for (int i = 0; i < N; i++) {
		cin >> A[i];
	}

	sort(A.rbegin(), A.rend());

	dp[X] = 1;

	for (int i = 0; i < N; i++) {

		for (int j = 0; j <= X; j++) {
			ndp[j] = (dp[j] * (N - i - 1)) % MOD;
		}
		for (int j = 0; j <= X; j++) {
			int t = j % A[i];
			ndp[t] += (dp[j]);
			if (ndp[t] >= MOD)ndp[t] -= MOD;
		}
		swap(dp, ndp);
	}


	for (int j = 0; j <= X; j++) {
		res += (j * dp[j]) % MOD;
		if (res >= MOD)res -= MOD;
	}

	cout << res << endl;
}