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>
using namespace std;

#define NDEBUG
#ifdef DEBUG
#include "../cout11.h"
#undef NDEBUG
#endif
#include <cassert>

#ifdef DEBUG
#define debug(...) fprintf(stderr, __VA_ARGS__)
#else
#define debug(...)
#endif

typedef long long ll;
typedef long double Double;
typedef unsigned long long ull;
typedef pair<int,int> ii;
typedef pair<ll,ll> llll;
typedef pair<double,double> dd;

typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<ii> vii;
typedef vector<vector<ii>> vvii;
typedef vector<ll> vll;
typedef vector<vector<ll>> vvll;
typedef vector<llll> vllll;
typedef vector<bool> vb;
typedef vector<string> vs;
typedef vector<double> vd;
typedef vector<long double> vD;

#define sz(a)  int((a).size())
#define pb  push_back
#define eb  emplace_back
#define FOR(var,from,to) for(int var=(from);var<=(to);++var)
#define rep(var,n)  for(int var=0;var<(n);++var)
#define rep1(var,n)  for(int var=1;var<=(n);++var)
#define repC2(vari,varj,n)  for(int vari=0;vari<(n)-1;++vari)for(int varj=vari+1;varj<(n);++varj)
#define repC3(vari,varj,vark,n)  for(int vari=0;vari<(n)-2;++vari)for(int varj=vari+1;varj<(n)-1;++varj)for(int vark=varj+1;vark<(n);++vark)
#define ALL(c)  (c).begin(),(c).end()
#define RALL(c)  (c).rbegin(),(c).rend()
#define tr(i,c)  for(auto i=(c).begin(); i!=(c).end(); ++i)
// #define found(s,e)  ((s).find(e)!=(s).end())
#define mset(arr,val)  memset(arr,val,sizeof(arr))
#define mid(x,y) ((x)+((y)-(x))/2)
#define IN(x,a,b) ((a)<=(x)&&(x)<=(b))
#define cons make_pair
#define clamp(v,lo,hi) min(max(v,lo),hi)

// #define found(s,e)  ((s).find(e)!=(s).end())
template <typename T>
inline bool found(set<T> &s, T elem) { return s.find(elem) != s.end(); }
template <typename T>
inline bool found(unordered_set<T> &s, T elem) { return s.find(elem) != s.end(); }
template <typename T, typename U>
inline bool found(map<T,U> &s, U elem) { return s.find(elem) != s.end(); }
template <typename T, typename U>
inline bool found(unordered_map<T,U> &s, U elem) { return s.find(elem) != s.end(); }
template <typename T>
inline bool found(vector<T> &s, T elem) { return find(s.begin(), s.end(), elem) != s.end(); }
// found(str, *) : returns indec, -1 if not found
inline int found(string &s, string t) { return (int)s.find(t); }
inline int found(string &s, int c) { return (int)s.find(c); }

template<typename T1, typename T2> inline void amin(T1 & a, T2 const & b) { if (a>b) a=b; }
template<typename T1, typename T2> inline void amax(T1 & a, T2 const & b) { if (a<b) a=b; }
template<typename X, typename T> auto vectors(X x, T a) { return vector<T>(x, a); }
template<typename X, typename Y, typename Z, typename... Zs> auto vectors(X x, Y y, Z z, Zs... zs) { auto cont = vectors(y, z, zs...); return vector<decltype(cont)>(x, cont); }

inline ll square(ll x) { return x * x; }
inline ll gcd(ll a, ll b) { while(a) swap(a, b%=a); return b; }
template <typename T>
inline T mod(T a, T b) { return ((a % b) + b) % b; }

template <typename T>
int find_left(vector<T>& v, T elem) {
    // elemそのものがあればそのindexを、なければ１つ下（左）のindexを返す
    return (int)(upper_bound(v.begin(), v.end(), elem) - v.begin()) - 1;
}
template <typename T>
int find_right(vector<T>& v, T elem) {
    // elemそのものがあればそのindexを、なければ１つ上（右）のindexを返す
    return (int)(lower_bound(v.begin(), v.end(), elem) - v.begin());
}

const ll MOD=1000000007LL;

ll ADD(ll x, ll y) { return (x+y) % MOD; }
ll SUB(ll x, ll y) { return (x-y+MOD) % MOD; }
ll MUL(ll x, ll y) { return x*y % MOD; }
ll POW(ll x, ll e) { ll v=1; for(; e; x=MUL(x,x), e>>=1) if (e&1) v = MUL(v,x); return v; }
ll DIV(ll x, ll y) { /*assert(y%MOD!=0);*/ return MUL(x, POW(y, MOD-2)); }
// ll comb(ll n, ll k) { ll v=1; for(ll i=1; i<=k; i++) v = DIV(MUL(v, n-i+1),i); return v; }
//

#define INTSPACE 12
char _buf[INTSPACE*1000000 + 3];

int loadint() {
    if (fgets(_buf, INTSPACE+3, stdin)==NULL) return 0;
    return atoi(_buf);
}

int loadvec(vector<int>& v, int N=-1) {
    if (N == 0) {
        v.clear();
        return 0;
    }
    if (N == -1) {
        N = loadint();
        if (N==0) return 0;
    }
    int bufsize = INTSPACE*N + 3;
    if (fgets(_buf, bufsize, stdin)==NULL) return 0;
    v.resize(N);

    int i=0;
    bool last = false;
    for (char *p=&_buf[0]; ;) {
        char *q = p;
        while (*q > ' ') ++q;
        if (*q == 0x0D || *q == 0x0A) last = true;
        *q = 0;
        v[i++] = atoi(p);
        if (last || i == N) break;
        p = q+1;
    }
    // assert(i <= N);
    return i;
}
void read_cr() {
    fgets(_buf, 256, stdin);
}

long double  solve(int N, vi& a) {
    if (N == 1) return a[0];

    sort(ALL(a));
    ll numer = a[0] + a[1];
    ll denom = 2;

    for (int i=2; i<N; ++i) {
        numer += denom * a[i];
        denom *= 2;
    }
    long double x = (long double)numer / (long double)denom;
    return x;
}

int main() {
    vi a;
    int N = loadvec(a);
    long double x = solve(N,a);
    printf("%.14Lf\n", x);
    return 0;
}