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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.*;
public class Main {
	
	public static int[] S;
	public static int m7 = (int) (Math.pow(10, 9) + 7);
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		long X = sc.nextLong();
		ArrayList<Long> set = new ArrayList<Long>();
		long min = Integer.MAX_VALUE;
		for(int i=0;i<N;i++) {
			long s = sc.nextLong();
			if (min > s) min = s;
			set.add(s);
		}
		System.out.println(func(N,set,min,X));
	}
	
	// index iと　最小値 minと割る数 Xを受け取る　X%iが最小値よりも小さいならば、 X%i*(N-i-1)!を返す
	// そうでない場合は、 i以降の数字jに対して再起した結果の和を返す
	public static long func(int N, ArrayList<Long> set, long min, long X) {
		if (set.size()==1) return X%set.get(0);
		long ret = 0;
		for(int idx = 0;idx<set.size();idx++) {
			if(X%set.get(idx)<min) {
				long cret = X%set.get(idx);
				for(int i=set.size()-1;i>0;i--) {
					cret = (cret * i) % m7;
					//System.out.println(cret);
				}
				ret = (ret + cret) % m7;
			}
			else {//X%set.get(idx)よりも大きいものでは悪い見ない
				ArrayList<Long> nset = new ArrayList(set);
				long s = X%set.get(idx);
				long cret = 1;
				long org = nset.size();
				nset.remove(idx);
				for(int i=0;i<nset.size();i++) {
					if(s < nset.get(i)) {
						
						
						cret = (cret*nset.size()) % m7;
						nset.remove(i);
						i--;
					}
				}
				
				ret = (ret + ((func(N,nset,min,X%set.get(idx))*cret)%m7)) % m7;
			}
		}
		
		return ret%m7;
	}
}