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: Wrong Answer)

import java.util.*;

class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int X = sc.nextInt();
		int[] S = new int[N];
		for (int i=0;i<N;i++) {
			S[i]=sc.nextInt();
		}

		Arrays.sort(S);
		double[] dp = new double[X+1];
		for (int i=1;i<=X;i++) {
			List<Integer> list = new ArrayList<Integer>();
			for (int j=0;j<N;j++) {
				if (S[j]<=i) {
					list.add(S[j]);
				}
			}
			if (list==null || list.size()==0) {
				dp[i]=i;
			}else {
				int n = list.size();
				double temp = 0;
				for (int j=0;j<n;j++) {
					temp=temp+dp[(i%list.get(j))]/n;
				}
				dp[i]=temp;
			}
		}

		long kaijo=1;
		for (int i=1;i<=N;i++) {
			kaijo=kaijo*i%1000000007;
		}
		System.out.println(dp[X]*kaijo%1000000007);
	}
}