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;
using ll = long long;

// double じゃ精度足りない
#define mod 1000000007
ll dp[201][100001];

int main() {
    int N, X;
    cin >> N >> X;
    vector<int>S(N);
    for (int i = 0; i < N; ++i) {
      cin >> S[i];
    }

    sort(S.begin(), S.end());
    // dp[残り回数][現在のX] = 期待値
    for (int i = 0; i <= X; ++i) {
        dp[0][i] = i;
    }
    for (int i = 1; i <= N; ++i) {
        // 1/i の確率で採択
        for (int j = 0; j <= X; ++j) {
            dp[i][j] = dp[i-1][j % S[i-1]];
            dp[i][j] += dp[i-1][j] * (i - 1);
            dp[i][j] %= mod;
            // 分母のiと最後のN!で相殺
        }
    }
    cout << dp[N][X];
    return 0;
}