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 <bits/stdc++.h>

using namespace std;

#define X first
#define Y second
#define mp make_pair
#define pb push_back
#define Debug(...) fprintf(stderr, __VA_ARGS__)

typedef long long LL;
typedef long double LD;
typedef unsigned int uint;
typedef pair <int, int> pii;
typedef unsigned long long uLL;

template <typename T> inline void Read(T &x) {
  char c = getchar();
  bool f = false;
  for (x = 0; !isdigit(c); c = getchar()) {
    if (c == '-') {
      f = true;
    }
  }
  for (; isdigit(c); c = getchar()) {
    x = x * 10 + c - '0';
  }
  if (f) {
    x = -x;
  }
}

template <typename T> inline bool CheckMax(T &a, const T &b) {
  return a < b ? a = b, true : false;
}

template <typename T> inline bool CheckMin(T &a, const T &b) {
  return a > b ? a = b, true : false;
}

const int N = 305;

int n, m, a[N], per[N];
LL sum, cnt, f[N][35];

inline LL Solve() {
  for (int i = 0; i <= n * m; ++i) {
    for (int j = 0; j < 1 << n - 1; ++j) {
      f[i][j] = 0;
    }
  }
  f[n * a[n - 1]][0] = 1;
  for (int i = 0; i < n * m; ++i) {
    for (int j = i; j <= n * m; ++j) {
      for (int k = 0; k < 1 << n - 1; ++k) {
        if (i % n && !(k >> per[i % n - 1] & 1)) {
          f[min(n * m, max(j, i + n * a[per[i % n - 1]]))][k | 1 << per[i % n - 1]] += f[j][k];
        }
      }
    }
  }
  return f[n * m][(1 << n - 1) - 1];
}

int main() {
#ifdef wxh010910
  freopen("d.in", "r", stdin);
#endif
  Read(n), Read(m);
  for (int i = 0; i < n; ++i) {
    Read(a[i]);
  }
  sort(a, a + n);
  for (int i = 0; i < n - 1; ++i) {
    per[i] = i;
  }
  do {
    sum += Solve();
    ++cnt;
  } while (next_permutation(per, per + n - 1));
  for (int i = 0; i < n - 1; ++i) {
    cnt = cnt * m;
  }
  printf("%.12lf\n", (double)sum / cnt);
#ifdef wxh010910
  Debug("My Time: %.3lfms\n", (double)clock() / CLOCKS_PER_SEC);
#endif
  return 0;
}