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 <stdio.h>
#include <algorithm>

using namespace std;

const long long mod = 1000000007;
long long n , x;
long long a[210];
long long dp[210][120000];
long long ans;
bool cmp ( long long x1 , long long x2 ) {
	return x1 > x2;
}
void work () {
	long long i , j;
	scanf ( "%lld%lld" , &n , &x );
	for ( i = 1 ; i <= n ; i++ ) {
		scanf ( "%lld" , &a[i] );
	}
	sort ( a + 1 , a + 1 + n , cmp );
	dp[0][x] = 1;
	for ( i = 0 ; i < n ; i++ ) {
		for ( j = 0 ; j <= x ; j++ ) {
			dp[i+1][j] = (dp[i+1][j] + dp[i][j] * (n-i-1)) % mod;
			dp[i+1][j%a[i+1]] = (dp[i+1][j%a[i+1]] + dp[i][j]) % mod;
		}
	}
	for ( i = 0 ; i <= x ; i++ ) {
		ans = (ans + dp[n][i] * i) % mod;
		//printf ( "%lld\n" , dp[n][i] );
	}
	printf ( "%lld\n" , ans );
}
int main () {
	work ();
	return 0;
}
