Problem: You have a circle of length C, and you are placing N arcs on it. Arc i has length L_i.

Every arc i is placed on the circle uniformly at random: a random real point on the circle is chosen, then an arc of length L_i centered at this point appears.

Note that the arcs are placed independently. For example, they may intersect or contain each other.

What is the probability that every real point of the circle will be covered by at least one arc? Assume that an arc covers its ends.

Constraints

* 2 \leq N \leq 6
* 2 \leq C \leq 50
* 1 \leq L_i < C
* All input values are integers.

Input

Input is given from Standard Input in the following format:


N C
L_1 L_2 ... L_N


Output

Print the probability that every real point of the circle will be covered by at least one arc. Your answer will be considered correct if its absolute error doesn't exceed 10^{-11}.

Examples

Input

2 3
2 2


Output

0.3333333333333333


Input

4 10
1 2 3 4


Output

0.0000000000000000


Input

4 2
1 1 1 1


Output

0.5000000000000000


Input

3 5
2 2 4


Output

0.4000000000000000


Input

4 6
4 1 3 2


Output

0.3148148148148148


Input

6 49
22 13 27 8 2 19


Output

0.2832340720702695

Solution (in CPP):

Solution:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <iomanip>
#include <algorithm>
#include <cstdlib>
#include <ctime>

using namespace std;

int n, c, a[6];
long long f[1 << 5][1 << 5][305];
int main() {
	scanf("%d %d", &n, &c);
	for (int i = 0; i < n; i++) cin >> a[i], a[i] *= n;
	c *= n;
	sort(a, a + n);
	f[0][0][a[n - 1]] = 1;
 // cout << a[n - 1] << endl;
	for (int x = 1; x < c; x++) if (x % n) {//x表示当前准备放的位置共用cn种可能
		for (int l = x + 1; l <= c; l++) {//l表示最右边覆盖到的位置
			for (int i = 0; i < 1 << n - 1; i++) if (!(i >> (x % n - 1) & 1)) { //找一个位置的小数部分状态为i，且和x对应的小数位置没有放
				for (int j = 0; j < 1 << n - 1; j++) if (f[i][j][l]){//找一个线段使用情况为j
					for (int y = 0; y < n - 1; y++) if (!(j >> y & 1)) {//选用第y条线段
            //cout << (i | 1 << (x % n - 1)) << " --- " << (j | 1 << y) << min(c, max(l, x + a[y])) << endl;
						f[i | 1 << (x % n - 1)][j | 1 << y][min(c, max(l, x + a[y]))] += f[i][j][l]; //更新结果
					}
				}
			}
		}
	}
	long long ans = f[(1 << n - 1) - 1][(1 << n - 1) - 1][c];
	long long all = 1;
	for (int i = 1; i < n; i++) {
		all = all * i * c / n;
	}
//	cout << ans << " " << all << endl;
	printf("%.13Lf\n",(long double) ans / all);


}