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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
#include <numeric>
using namespace std;
using ll = long long;
const ll mod =  1000000007;

#define REP(i, n) for (int i = 0; i < (n); ++i)
#define REP_FROM(i, j, n) for (int i = (j); i < (n); ++i)
#define all(x) (x).begin(),(x).end()

ll power(ll base, ll exponent, ll module){ if(exponent % 2){ return power(base, exponent - 1, module) * base % module; }else if(exponent){ ll root_ans = power(base, exponent / 2, module); return root_ans * root_ans % module; }else{ return 1; }}
ll inverse(ll x){ return power(x, mod - 2, mod);}
struct combination{ vector<ll> fact, inv; combination(int sz) : fact(sz + 1), inv(sz + 1){ fact[0] = 1; for(int i = 1; i <= sz; i++){ fact[i] = fact[i - 1] * i % mod; } inv[sz] = power(fact[sz], mod - 2, mod); for(int i = sz - 1; i >= 0; i--){ inv[i] = inv[i + 1] * (i + 1) % mod; } } ll C(int p, int q) const{ if(q < 0 || p < q) return 0; return (fact[p] * inv[q] % mod * inv[p - q] % mod); } };

signed main() {
  ios::sync_with_stdio(false);
  int n, x;
  cin >> n >> x;
  vector<ll> s(n);
  REP(i, n) cin >> s[i];
  sort(all(s), greater<ll>());
  vector<vector<ll>> dp(n+1, vector<ll>((x+1)));
  dp[0][x] = 1;
  REP(i, n) {
    REP(j, x+1) {
      dp[i+1][j%s[i]] += dp[i][j];
      dp[i+1][j%s[i]] %= mod;
      dp[i+1][j] += dp[i][j] * (n-i-1);
      dp[i+1][j] %= mod;
    }
  }
  ll ans = 0;
  REP(i, x) {
    ans += dp[n][i] * i;
    ans %= mod;
  }
  cout << ans << endl;
}