Problem: Esports is a form of competitive sports using video games. Dota 2 is one of the most popular competitive video games in Esports. Recently, a new video game Dota 3 was released. In Dota 3 a player can buy some relics for their hero. Relics are counters that track hero's actions and statistics in a game.

Gloria likes to play Dota 3, so she wants to buy all n available relics for her favorite hero.

Relics can be bought using an in-game currency called shards. Each relic has its own price — c_i shards for the i-th relic. A player can buy a relic using one of the following options: 

  * Pay c_i shards to buy the i-th relic; 
  * Pay x shards and randomly get one of all n relics. The probability of getting a relic is the same for all n relics. If a duplicate relic is received, then the relic is recycled and x/2 shards are given back to the player. 



Gloria wants to buy all n relics. Help her minimize the expected number of shards she spends to buy all the relics.

Input

The first line contains two integers n and x (1 ≤ n ≤ 100; 1 ≤ x ≤ 10 000) — the number of relics and the cost to receive a random relic.

The second line consists of n integers c_1, c_2, …, c_n (x ≤ c_i ≤ 10 000; ∑{c_i} ≤ 10 000) — the prices of n relics.

Output

Print a single real number — the minimum expected number of shards that Gloria must spend to buy all the relics.

The absolute or relative error should not exceed 10^{-9}.

Examples

Input


2 20
25 100


Output


47.50000000000000000


Input


4 30
60 50 60 80


Output


171.25000000000000000

Note

In the first example, the optimal strategy is to randomly get one of the two relics paying 20 shards. Then there are two scenarios. 

The first one happens if Gloria receives the first relic. Then she keeps getting random relics until she obtains the second relic. The expected number of shards to spend in this scenario is 20 + 30 = 50.

In the second scenario, Gloria initially gets the second relic. Then it is better to buy the first relic for 25 shards, so the expected number of shards to spend in this scenario is 20 + 25 = 45.

Thus, the expected number of shards to spend is (50 + 45)/(2) = 47.5.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MOD = (int)1e9 + 7;
const int FFTMOD = 119 << 23 | 1;
const int INF = (int)1e9 + 23111992;
const long long LINF = (long long)1e18 + 23111992;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
inline long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long fpow(long long n, long long k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
template <class T>
inline int chkmin(T& a, const T& val) {
  return val < a ? a = val, 1 : 0;
}
template <class T>
inline int chkmax(T& a, const T& val) {
  return a < val ? a = val, 1 : 0;
}
inline unsigned long long isqrt(unsigned long long k) {
  unsigned long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
inline int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
inline int sign(long double x) { return x < -EPS ? -1 : x > +EPS; }
inline int sign(long double x, long double y) { return sign(x - y); }
mt19937 mt(chrono::high_resolution_clock::now().time_since_epoch().count());
inline int mrand() { return abs((int)mt()); }
void chemthan() {
  int n, x;
  cin >> n >> x;
  vector<int> c(n);
  for (int i = (0); i < (n); ++i) cin >> c[i], c[i] <<= 1, c[i] -= x;
  sort((c).begin(), (c).end());
  int K = accumulate((c).begin(), (c).end(), 0) + 1;
  double res = 0;
  auto ff = [&](int k) { return (double)x * (n + n - k) / (n - k); };
  vector<vector<double>> dp(n + 1, vector<double>(K));
  vector<vector<double>> ndp(n + 1, vector<double>(K));
  for (auto& r : dp)
    for (auto& e : r) e = 0;
  dp[0][0] = 1;
  for (int i = (0); i < (n); ++i) {
    for (auto& r : ndp)
      for (auto& e : r) e = 0;
    for (int j = (0); j < (i + 1); ++j) {
      for (int k = (0); k < (K); ++k) {
        if (k + c[i] < K) {
          ndp[j + 1][k + c[i]] += dp[j][k] / (n - j) * (j + 1);
        }
        ndp[j][k] += dp[j][k];
      }
    }
    swap(dp, ndp);
  }
  for (int i = (0); i < (n); ++i) {
    for (int k = (0); k < (K); ++k)
      if (K - 1 - k >= n * x) {
        res -= dp[i][k] * ((double)(K - 1 - k) / (n - i) + x - ff(i));
      }
  }
  for (int i = (0); i < (n); ++i) {
    res += c[i] + x;
  }
  cout << fixed << setprecision(9) << res / 2 << "\n";
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(0), cin.tie(0);
  if (argc > 1) {
    assert(freopen(argv[1], "r", stdin));
  }
  if (argc > 2) {
    assert(freopen(argv[2], "wb", stdout));
  }
  chemthan();
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}