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;
#define ll long long
#define all(V) V.begin(),V.end()
const ll MOD = 1000000007;

vector<int> G[100010];
ll fuhen[100010];
ll arr[100010];

int main(){
	int N, X;
	cin >> N >> X;
	vector<int> S(N);
	for (int i = 0;i < N;i++) {
		cin >> S[i];
	}
	for (int i = 0;i <= X;i++) {
		for (int j = 0;j < N;j++) {
			if (i < S[j]) {
				fuhen[i]++;
			}
			else {
				G[i].push_back(i % S[j]);
			}
		}
	}
	arr[X] = 1;
	for (ll i = 0;i < N;i++) {
		for (int j = 0;j <= X;j++) {
			if (arr[j] == 0)continue;
			for (int x : G[j]) {
				arr[x] += arr[j];
				arr[x] %= MOD;
			}
			if (fuhen[j] > i) arr[j] *= (fuhen[j] - i);
			else arr[j] = 0;
			arr[j] %= MOD;
		}
	}
	ll ans = 0;
	for (ll i = 0;i <= X;i++) {
		ans += arr[i] * i;
		ans %= MOD;
	}
	cout << ans << endl;
}