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 be(v) (v).begin(),(v).end()
#define pb(q) push_back(q)
typedef long long ll;
using namespace std;
const ll mod=1000000007;
#define doublecout(a) cout<<fixed<<setprecision(10)<<a<<endl;

int main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(false);
    ll n,x;
    cin>>n>>x;
    set<ll> s={x};
    vector<ll> a(n),dp(x+1,0LL);
    for(int i=0;i<n;i++)cin>>a[i];dp[x]=1;
    sort(a.rbegin(),a.rend());
    for(int i=0;i<n;i++){
    	ll m=a[i];
    	for(auto& v:s){
    		s.insert(v%m);
    		ll k=dp[v];
    		(dp[v]*=(n-i-1))%=mod;
    		(dp[v%m]+=k)%=mod;
    	}
    }
    ll ans=0;
    for(int i=0;i<=x;i++)(ans+=dp[i]*i)%=mod;
    cout << ans << endl;
    return 0;
}
