Problem:
Score : 300 points

Problem Statement

You have a pot and N ingredients. Each ingredient has a real number parameter called value, and the value of the i-th ingredient (1 \leq i \leq N) is v_i.

When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.

After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Constraints

2 \leq N \leq 50

1 \leq v_i \leq 1000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
v_1 v_2 \ldots v_N

Output

Print a decimal number (or an integer) representing the maximum possible value of the last ingredient remaining.

Your output will be judged correct when its absolute or relative error from the judge's output is at most 10^{-5}.

Sample Input 1

2
3 4

Sample Output 1

3.5

If you start with two ingredients, the only choice is to put both of them in the pot. The value of the ingredient resulting from the ingredients of values 3 and 4 is (3 + 4) / 2 = 3.5.

Printing 3.50001, 3.49999, and so on will also be accepted.

Sample Input 2

3
500 300 200

Sample Output 2

375

You start with three ingredients this time, and you can choose what to use in the first composition. There are three possible choices:

Use the ingredients of values 500 and 300 to produce an ingredient of value (500 + 300) / 2 = 400. The next composition will use this ingredient and the ingredient of value 200, resulting in an ingredient of value (400 + 200) / 2 = 300.

Use the ingredients of values 500 and 200 to produce an ingredient of value (500 + 200) / 2 = 350. The next composition will use this ingredient and the ingredient of value 300, resulting in an ingredient of value (350 + 300) / 2 = 325.

Use the ingredients of values 300 and 200 to produce an ingredient of value (300 + 200) / 2 = 250. The next composition will use this ingredient and the ingredient of value 500, resulting in an ingredient of value (250 + 500) / 2 = 375.

Thus, the maximum possible value of the last ingredient remaining is 375.

Printing 375.0 and so on will also be accepted.

Sample Input 3

5
138 138 138 138 138

Sample Output 3

138

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <iostream>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <set>
#include <map>
#include <random>
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define repp(i,n,m) for (int i = m; i < (n); ++i)
#define repl(i,n) for (long long i = 0; i < (n); ++i)
#define reppl(i,n,m) for (long long i = m; i < (n); ++i)
using namespace std;
using ll = long long;
using ld = long double;
using P = pair<int, int>;
using PL = pair<long long, long long>;
using Pxy = pair<double, double>;
using T = tuple<int, int, bool>;
const int INF = 1001001007;
const int mod = 1000000007;
//const int MOD = 998244353;
const ll inf = 1e18;
template <typename AT>
void printvec(vector<AT> &ar){
    rep(i,ar.size()-1) cout << ar[i] << " ";
    cout << ar[ar.size()-1] << endl;
}
template <typename Q>
void printvvec(vector<vector<Q>> &ar){
    rep(i,ar.size()){
        rep(j,ar[0].size()-1) cout << ar[i][j] << " ";
        cout << ar[i][ar[0].size()-1] << endl;
    }
}
template <typename S>
bool range(S a, S b, S x){
    return (a <= x && x < b);
}

struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const {
    mint res(*this);
    return res+=a;
  }
  mint operator-(const mint a) const {
    mint res(*this);
    return res-=a;
  }
  mint operator*(const mint a) const {
    mint res(*this);
    return res*=a;
  }
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const {
    return pow(mod-2);
  }
  mint& operator/=(const mint a) {
    return (*this) *= a.inv();
  }
  mint operator/(const mint a) const {
    mint res(*this);
    return res/=a;
  }
};

mint f(int n) {
  if (n == 0) return 1;
  mint x = f(n/2);
  x *= x;
  if (n%2 == 1) x *= 2;
  return x;
}

mint choose(int n, int a) {
  mint x = 1, y = 1;
  rep(i,a) {
    x *= n-i;
    y *= i+1;
  }
  return x / y;
}


int main() {
    int n; cin >> n;
    vector<ld> ar(n);
    ld ans = 0;
    rep(i,n) cin >> ar[i];
    sort(ar.begin(),ar.end());
    ans = (ar[0] + ar[1]) / 2.0;
    repp(i,n,2) ans = (ans + ar[i]) / 2.0;
    cout << setprecision(20);
    cout << ans << endl;
}