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;

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef pair<int, int> ii;
typedef pair<ll, int> li;
typedef pair<int, ll> il;
typedef vector<ii> vii;
typedef vector<il> vil;
typedef vector<li> vli;

#define F            first
#define S            second
#define pb(x)        push_back(x)
#define mp(a,b)      make_pair(a,b)
#define mt(a,b,c)    make_tuple(a,b,c);
#define sz           size()
#define all(A)       A.begin(), A.end()
#define mem(a, b)    memset(a, b, sizeof(a))   // all elements of A are b=0 or b=-1 else don't work
#define f0(i,b)      for(ll i=0;i<(b);i++)
#define f1(i,b)      for(ll i=1;i<=(b);i++)
#define f2(i,a,b)    for(ll i=(a);i<=(b);i++)
#define f3(i,b,a)    for(ll i=(b);i>=(a);i--)
#define rep(i,a,b,c) for(ll i=(a);i!=(b);i+=(c))
#define mod          1000000007

#define whilee(i,t)  while(i!=t)
#define  sl(a)       scanf("%lld",&a)
#define  pl(a)       printf("%lld\n",a)

#define scan(x)      do{while((x=getchar())<'0'); for(x-='0'; '0'<=(_=getchar()); x=(x<<3)+(x<<1)+_-'0');}while(0)
char _;

//for(auto it =A.begin(); it!=A.end(); it++)
//sort(A.begin(),A.end()); //sort(A, +n, greater<int>());
//binary_search(A.begin(),A.end(),x);
//reverse(A.begin(), A.end());
//cout << *max_element(A.begin(), A.end());
//cout << *min_element(A.begin(), A.end());
//count(A.begin(), A.end(), X); //counts the occurrences of X
//distance(A.begin(),A.end());  //distance between first to last element
//accumulate(A.begin(), A.end(), 0); //add
//next_permutation(A.begin(), A.end());
//prev_permutation(A.begin(), A.end());
//swap(S[0], T[0]);

//scanf("%lld",&x);
//printf("%lld\n",x);
//while (scanf("%lld %lld", &a, &b) == 2)
//printf("Case %lld: %lld\n", t++, a);

//vector <ll> A;
//deque <ll> A;
//queue <ll> A;
//stack <ll> A;
//list <int> A;
//set <ll> A;
//map <ll,ll> A;
//pair <ll,ll> A ;
//tuple <ll, ll, ll> A;
//vector <pair <ll,ll>> A ;
//bitset<32> bset1;
//bool ok = true;
//char ch;
//string S, T, U;
ll i=0, j=0, k=0, a, b, c, d, e, f, g, h, l, m, n, o, p, q, r, s, t=1, u, v, w, maxx=INT_MIN, minn=INT_MAX, ans=0, sum=0, cnt=0;
double x=56, y, z;
int main()
{

    //scan(n);
    cin>>n;
    double A[n];
    f0(i,n)
    {
        cin>>A[i];
    }
    sort(A,A+n);
    for(ll i=1;i<(n);i++)
    {
        x = (A[i-1]+A[i])/2;
        A[i] = x;
    }
    printf("%.5f\n",x);
    return 0;
}