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 <iostream>
#include <vector>
#include <algorithm>

using namespace std;
typedef long long ll;
typedef vector<int> vi;
#define REP(i,n) for (ll i = 0; i < n; i++)
#define REPR(i,n) for (ll i = n-1; i >= 0; i--)
#define FOR(i,m,n) for (ll i = m; i < n; i++)
#define FORE(x,xs) for (auto& x : xs)
#define ALL(x) x.begin(), x.end()

const int MAX_N = 210;
const int MAX_X = 1e5+10;
const int MOD = 1e9+7;
ll dp[MAX_N][MAX_X];

int N, X;
vi S;

ll dfs (int n, int x) {
    if (dp[n][x] != -1) return dp[n][x];
    if (n==0) return dp[n][x] = x % S[0];
    return dp[n][x] = (n * dfs(n-1, x) + dfs(n-1, x%S[n])) % MOD;
}

int main() {
    cin >> N >> X;
    REP (n, MAX_N) REP (x, MAX_X) dp[n][x] = -1;
    S = vi(N);
    REP (i, N) cin >> S[i];
    sort(ALL(S));

    cout << dfs(N-1, X) << endl;
}