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: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<stdint.h>
#include<inttypes.h>

typedef int32_t i32;
typedef int64_t i64;

int cmp (const void *a, const void *b) {
  i32 d = *(i32 *)a - *(i32 *)b;
  return d == 0 ? 0 : d < 0 ? -1 : 1;
}

void run (void) {
  i32 n, x;
  scanf ("%" SCNi32 "%" SCNi32, &n, &x);
  i32 *a = (i32 *) calloc (n, sizeof (i32));
  for (i32 i = 0; i < n; ++i) {
    scanf ("%" SCNi32, a + i);
  }
  qsort (a, n, sizeof (i32), cmp);
  i32 *dp = (i32 *) calloc (x + 1, sizeof (i32));
  for (i32 i = 0; i <= x; ++i) {
    dp[i] = i;
  }
  const i32 mod = 1000000007;
  for (i32 i = 0; i < n; ++i) {
    for (i32 j = x; j >= 0; --j) {
      dp[j] = ((i64) i * dp[j] + dp[j % a[i]]) % mod;
    }
  }
  printf ("%" PRIi32 "\n", dp[x]);
}

int main (void) {
  run ();
  return 0;
}