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 = 1e9 + 7;
const int MX = 2e5 + 5;
const long long INF = 1e18;
const long double PI = 4 * atan((long double)1);
template <class T>
bool ckmin(T& a, const T& b) {
  return a > b ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
namespace input {
template <class T>
void re(complex<T>& x);
template <class T1, class T2>
void re(pair<T1, T2>& p);
template <class T>
void re(vector<T>& a);
template <class T, size_t SZ>
void re(array<T, SZ>& a);
template <class T>
void re(T& x) {
  cin >> x;
}
void re(double& x) {
  string t;
  re(t);
  x = stod(t);
}
void re(long double& x) {
  string t;
  re(t);
  x = stold(t);
}
template <class T, class... Ts>
void re(T& t, Ts&... ts) {
  re(t);
  re(ts...);
}
template <class T>
void re(complex<T>& x) {
  T a, b;
  re(a, b);
  x = cd(a, b);
}
template <class T1, class T2>
void re(pair<T1, T2>& p) {
  re(p.first, p.second);
}
template <class T>
void re(vector<T>& a) {
  for (int i = (0); i < ((int)a.size()); ++i) re(a[i]);
}
template <class T, size_t SZ>
void re(array<T, SZ>& a) {
  for (int i = (0); i < (SZ); ++i) re(a[i]);
}
}  // namespace input
using namespace input;
namespace output {
void pr(int x) { cout << x; }
void pr(long x) { cout << x; }
void pr(long long x) { cout << x; }
void pr(unsigned x) { cout << x; }
void pr(unsigned long x) { cout << x; }
void pr(unsigned long long x) { cout << x; }
void pr(float x) { cout << x; }
void pr(double x) { cout << x; }
void pr(long double x) { cout << x; }
void pr(char x) { cout << x; }
void pr(const char* x) { cout << x; }
void pr(const string& x) { cout << x; }
void pr(bool x) { pr(x ? "true" : "false"); }
template <class T>
void pr(const complex<T>& x) {
  cout << x;
}
template <class T1, class T2>
void pr(const pair<T1, T2>& x);
template <class T>
void pr(const T& x);
template <class T, class... Ts>
void pr(const T& t, const Ts&... ts) {
  pr(t);
  pr(ts...);
}
template <class T1, class T2>
void pr(const pair<T1, T2>& x) {
  pr("{", x.first, ", ", x.second, "}");
}
template <class T>
void pr(const T& x) {
  pr("{");
  bool fst = 1;
  for (const auto& a : x) pr(!fst ? ", " : "", a), fst = 0;
  pr("}");
}
void ps() { pr("\n"); }
template <class T, class... Ts>
void ps(const T& t, const Ts&... ts) {
  pr(t);
  if (sizeof...(ts)) pr(" ");
  ps(ts...);
}
void pc() { pr("]\n"); }
template <class T, class... Ts>
void pc(const T& t, const Ts&... ts) {
  pr(t);
  if (sizeof...(ts)) pr(", ");
  pc(ts...);
}
}  // namespace output
using namespace output;
namespace io {
void setIn(string second) { freopen(second.c_str(), "r", stdin); }
void setOut(string second) { freopen(second.c_str(), "w", stdout); }
void setIO(string second = "") {
  cin.sync_with_stdio(0);
  cin.tie(0);
  if ((int)second.size()) {
    setIn(second + ".in"), setOut(second + ".out");
  }
}
}  // namespace io
using namespace io;
typedef decay<decltype(MOD)>::type T;
struct mi {
  T val;
  explicit operator T() const { return val; }
  mi() { val = 0; }
  mi(const long long& v) {
    val = (-MOD <= v && v <= MOD) ? v : v % MOD;
    if (val < 0) val += MOD;
  }
  friend void pr(const mi& a) { pr(a.val); }
  friend void re(mi& a) {
    long long x;
    re(x);
    a = mi(x);
  }
  friend bool operator==(const mi& a, const mi& b) { return a.val == b.val; }
  friend bool operator!=(const mi& a, const mi& b) { return !(a == b); }
  friend bool operator<(const mi& a, const mi& b) { return a.val < b.val; }
  mi operator-() const { return mi(-val); }
  mi& operator+=(const mi& m) {
    if ((val += m.val) >= MOD) val -= MOD;
    return *this;
  }
  mi& operator-=(const mi& m) {
    if ((val -= m.val) < 0) val += MOD;
    return *this;
  }
  mi& operator*=(const mi& m) {
    val = (long long)val * m.val % MOD;
    return *this;
  }
  friend mi pow(mi a, long long p) {
    mi ans = 1;
    assert(p >= 0);
    for (; p; p /= 2, a *= a)
      if (p & 1) ans *= a;
    return ans;
  }
  friend mi inv(const mi& a) {
    assert(a != 0);
    return pow(a, MOD - 2);
  }
  mi& operator/=(const mi& m) { return (*this) *= inv(m); }
  friend mi operator+(mi a, const mi& b) { return a += b; }
  friend mi operator-(mi a, const mi& b) { return a -= b; }
  friend mi operator*(mi a, const mi& b) { return a *= b; }
  friend mi operator/(mi a, const mi& b) { return a /= b; }
};
int n, x;
vector<int> c;
long double w[101][10001];
int sum = 0;
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  re(n, x);
  c.resize(n);
  re(c);
  for (auto& t : c) sum += t;
  w[0][0] = 1;
  for (auto& t : c)
    for (int i = (n)-1; i >= (0); --i)
      for (int j = (0); j < (10001 - t); ++j) {
        w[i + 1][j + t] += w[i][j] * (i + 1) / (n - i);
      }
  long double ans = 0;
  for (int i = (0); i < (n); ++i)
    for (int j = (0); j < (10001); ++j) {
      ans += w[i][j] * min((long double)(sum - j) / (n - i),
                           ((long double)n / (n - i) + 1) * x / 2);
    }
  cout << fixed << setprecision(12);
  ps(ans);
}