Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

2 19
3 7

Sample Output 1

3

There are two possible orders in which we remove the numbers from S.

If we remove 3 and 7 in this order, the number on the blackboard changes as follows: 19 \rightarrow 1 \rightarrow 1.

If we remove 7 and 3 in this order, the number on the blackboard changes as follows: 19 \rightarrow 5 \rightarrow 2.

The output should be the sum of these: 3.

Sample Input 2

5 82
22 11 6 5 13

Sample Output 2

288

Sample Input 3

10 100000
50000 50001 50002 50003 50004 50005 50006 50007 50008 50009

Sample Output 3

279669259

Be sure to compute the sum modulo 10^{9}+7.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <stdio.h>
int compare(const void *a,const void *b){
    if(*(long long *)a>*(long long *)b)return 1;
    return -1;
}
int main()
{   int n,x;
    scanf("%d%d",&n,&x);
    int s[n];
    for(int i=0;i<n;i++){
    scanf("%d",&s[i]);
    }
    qsort(s,n,sizeof(int),compare);
    
    long long preans[200][200]={0};
    for(int i=n-1;i>=0;i--){
        for(int j=0;j<i;j++){
            int cpy=x;
            cpy%=s[j];
            preans[i][j]=cpy%s[i];
            //printf("preans[%d][%d]=%lld",i,j,preans[i][j]);
        }
        preans[i][i]=x%s[i];
        printf("\n");
    }
    long long ans=0;
    for(int i=n-1;i>=0;i--){
        for(int j=0;j<=i;j++){
            if(j>0){
                int k=j;
                while(k>0){
                    preans[i][j]*=k;
                    preans[i][j]&=1000000007;
                    k--;
                }
            }
            if(i<n-1){
                int k=i;
                while((n-1)-k>0){
                    preans[i][j]*=((n-1)-k);
                    preans[i][j]%=1000000007;
                    k++;
                }
            }
        }
    }
        for(int i=n-1;i>=0;i--){
            for(int j=0;j<i;j++){   
        ans+=preans[i][j];
        ans%=1000000007;
        //printf("ans=%d\n",ans);
        }
    }
    printf("%lld",ans);
    return 0;
}