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 <string>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
using ll = long long;

constexpr int P = 1000000007;

template <class S, class DP>
int f(int n, int x, const S& s, DP& dp) {
    if (n == 1) return x % s[n - 1];
    if (dp[n][x] >= 0) return dp[n][x];

    ll r = 0;
    r += f(n - 1, x % s[n - 1], s, dp);
    r += f(n - 1, x, s, dp) * (ll)(n - 1);
    return dp[n][x] = r % P;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    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());

    vector<vector<int>> dp(n + 1, vector<int>(x + 1, -1));

    cout << f(n, x, s, dp) << endl;

    return 0;
}