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.*;
public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int num = sc.nextInt();
		int start = sc.nextInt();
		long div[] = new long[num];
		for(int i = 0; i < num; i ++) {
			div[i] = sc.nextLong();
		}
		Arrays.sort(div);
		double dp[][] = new double[num][start + 1];
			for(int j = 0; j <= start; j ++) {
				dp[num - 1][j] = j % div[0];
		}

		for(int i = num - 2; i >= 0; i --) {
			for(int j = 0; j <= start; j ++) {
				dp[i][j] = 1.0 / (num - i) * dp[i + 1][(int)(j % div[num - i - 1])]
					+ (num - i - 1.0) / (num - i) * dp[i + 1][j];
			}
		}
		
		double ans = dp[0][start];
		for(int i = 1; i <= num; i ++) {
			ans = ans * i;
			if(ans % 1 == 0) {
				ans = mod((long)ans);
			}
		}
		System.out.println((long)ans);
	}

	static long divisor = (long)Math.pow(10, 9) + 7;
	public static long mod(long i) {
		return i % divisor + ((i % divisor) < 0 ? divisor : 0);
	}
}