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)

#ifndef BZ
#pragma GCC optimize "-O3"
#endif
#include <bits/stdc++.h>

#define FASTIO
#define ALL(v) (v).begin(), (v).end()
#define rep(i, l, r) for (int i = (l); i < (r); ++i)

#ifdef FASTIO
#define scanf abacaba
#define printf abacaba
#endif

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

using namespace std;

template<typename T> T mo(T x, T y) { x %= y; return x <= 0 ? x + y : x; }

const int MOD = 1000 * 1000 * 1000 + 7;

void ad(int& x, int y) {
    x += y;
    if (x >= MOD) {
        x -= MOD;
    }
}

int mul(int x, int y) {
    return (1ll * x * y) % MOD;
}

const int MX = 207;
const int MS = 100 * 1000 + 7;

int f[MX][MS];

int a[MX];

int main() {
#ifdef FASTIO
    ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#endif
    int n, x;
    cin >> n >> x;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    sort(a, a + n);
    reverse(a, a + n);
    f[0][x] = 1;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j <= x; j++) {
            if (j < a[i]) {
                ad(f[i + 1][j], mul(f[i][j], n - i));
            } else {
                ad(f[i + 1][j], mul(f[i][j], n - i - 1));
            }
            if (j >= a[i]) {
                ad(f[i + 1][j % a[i]], f[i][j]);
            }
        }
    }

    int ans = 0;
    for (int i = 0; i <= x; i++) {
        ad(ans, mul(i, f[n][i]));
    }

    cout << ans << "\n";
    return 0;
}