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 <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <vector>
#include <map>
#include <set>
#define MAXN 205
#define P 1000000007
#define G 3
#define INF 0x3f3f3f3f
#define rint register int
#define LL long long
#define LD long double
#define fuck printf("FUCK!!\n");
using namespace std;

int n, x, s[MAXN], f[MAXN][100005];

inline int add(int x, int y) {return x+y>=P?x+y-P:x+y;}
inline int sub(int x, int y) {return x>=y?x-y:x-y+P;}

int main()
{
    scanf("%d%d",  &n, &x);
    for(rint i=1; i<=n; ++i) scanf("%d", &s[i]);
    sort(s+1, s+n+1);
    for(rint i=1; i<=x; ++i) f[0][i]=i;
    for(rint i=1; i<=n; ++i)
        for(rint j=1; j<=x; ++j)
            f[i][j]=add(f[i-1][j%s[i]], 1LL*(i-1)*f[i-1][j]%P);
    printf("%d\n", f[n][x]);
    return 0;
}