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 <bits/stdc++.h>

#define all(A) begin(A), end(A)
#define sz(A) int(A.size())

using namespace std;

typedef long long ll;

const int MOD = 1e9 + 7, MAX_N = 210, MAX_VAL = 1e5 + 10;

inline int add (ll a, ll b) { return (a + b) % MOD; }
inline int mul (ll a, ll b) { return (a * b) % MOD; }

int n, x;
int dp[MAX_N][MAX_VAL];
vector <int> S;

int rec (int pos, int rem) {
  if (pos == n) return rem;
  int& ret = dp[pos][rem];
  if (ret != -1) return ret;
  ret = 0;
  ret = add(ret, rec(pos + 1, rem % S[pos]));
  ret = add(ret, mul(n - pos - 1, rec(pos + 1, rem)));
  return ret;
}

int main () {
  ios::sync_with_stdio(false); cin.tie(0);
  cin >> n >> x;
  S.resize(n);
  for (int i = 0; i < n; i++) cin >> S[i];
  sort(all(S));
  reverse(all(S));
  memset(dp, -1, sizeof dp);
  cout << rec(0, x) << endl;
  return (0);
}
