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 <stdio.h>
#include <array>
#include <algorithm>
#include <vector>

constexpr int MAX_X = 1e5;
constexpr int MAX_N = 200;
constexpr int mod = 1e9 + 7;
std::vector<int> S;
std::array<std::array<long long int, MAX_X+1>, MAX_N> DP;

int main(void){
    int n, X; scanf("%d%d", &n, &X);
    S.resize(n); for(int i = 0; i < n; ++i) scanf("%d", &S[i]);
    // DP[n][x] := 数字 x を s0 ... sn-1 で やった時の解
    // 遷移式 : DP[i][0] = 0 for all i 
    //       : DP[0][x] = x % s0
    //         DP[i][x] = DP[i-1][x] * i + DP[i-1][x % i]
    std::sort(S.begin(), S.end());
    for(int x = 0; x <= X; ++x) DP[0][x] = x % S[0];
    for(int i = 0; i < n; ++i) DP[i][0] = 0;
    for(int i = 1; i < n; ++i){
        for(int x = 0; x <= X; ++x) DP[i][x] = (DP[i-1][x] * i + DP[i-1][x % S[i]]) % mod;
    }
    printf("%lld\n", DP[n-1][X]);
    return 0;
}