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;
typedef long long ll;
const ll MOD = ll(1e9+7);
#define REP(i, n) for (int i = 0; i < n; i++)

ll n, x, s[205], dp[205][100005];  // i番目の数までみて値をjにする通り

int main() {
    cin >> n >> x;
    REP(i,n){
        cin >> s[i];
    }
    // 降順にみる。 a<b ならば x%a%b = x%a よりbを無視できるため
    sort(s,s+n);
    reverse(s,s+n);

    dp[0][x] = 1;
    REP(i,n){
        REP(j,x+1){
            // 使う遷移
            dp[i+1][j%s[i]] += dp[i][j];
            dp[i+1][j%s[i]] %= MOD;
            // 使わない遷移
            // これより後のどこかで使う
            dp[i+1][j] += dp[i][j] * (n-i-1);
            dp[i+1][j] %= MOD;
        }
    }
    ll ans = 0;
    REP(i,x+1){
        ans += dp[n][i] * i % MOD;
        ans %= MOD;
    }
    cout << ans << endl;
    return 0;
}