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>
using namespace std;

long long n, x, dp[200000], dpw[200000];
vector<long long> vec;

long long pw (long long a, long long b)
{
	long long ret=1;
	for (;b;b >>= 1, a=1ll * a * a % 1000000007)
		if (b & 1)
			ret=1ll * ret * a % 1000000007;
	return ret;
}

int main ()
{
	cin>>n>>x;
	long long cur=1;
	for (int i=0;i < n;i++)
	{
		int a;
		cin>>a;
		if (a<=x)
			vec.push_back(a);
		cur=(cur * (i + 1)) % 1000000007;
	}
	sort(vec.begin(), vec.end());
	for (int i=0;i<=n;i++)
		dpw[i]=pw(i, 1000000007 - 2);
	for (int i=0;i<=x;i++)
	{
		int pos=upper_bound(vec.begin(), vec.end(), i) - vec.begin();
		if (pos==0)
			dp[i]=i;
		for (int j=0;j < pos;j++)
			dp[i]=(dp[i] + dpw[pos] * dp[i % vec[j]]) % 1000000007;
	}
	cout << 1ll * dp[x] * cur % 1000000007 << endl;
}