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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>

#define NEW(p,n){p=malloc((n)*sizeof(p[0]));if(p==NULL){printf("not enough memory\n");exit(1);};}
//pの型の変数n個の要素分のメモリを確保し、そのアドレスをpに代入するマクロ


//ソートされたA[left~mid]とA[mid+1~right]をくっつける関数
void merge(double* A, int left, int mid, int right, double* B){
    int i,j,k;
    double tmp;

    //kは一時的な配列Bの現在の添字、iは前半A[left~mid]、jは後半A[mid+1~right]を指す添字
    k=left; i=left; j=mid+1;
    while(k<=right){
        if(j>right) tmp=A[i++]; //前半のみにデータがある
        else if(i>mid) tmp=A[j++]; //後半のみにデータがある
        else if(A[i]<=A[j]) tmp=A[i++]; //前半の方が小さい
        else tmp=A[j++]; //後半の方が小さい

        B[k++]=tmp;
    }
    for(i=left;i<=right;i++) A[i]=B[i]; //元の配列に戻す
}

//A[left]~A[right]までの配列を昇順にマージソートする関数
void merge_sort(double* A, int left, int right, double*B){
    int mid;

    if(left<right){
        mid=(left+right)/2;
        merge_sort(A,left,mid,B); //A[left]~A[mid]までを再帰的にソート
        merge_sort(A,mid+1,right,B); //A[mid+1]~A[right]までを再帰的にソート
        merge(A,left,mid,right,B); //分割した二つをくっつける
    }
}

int main(void){
    int N;
    scanf("%d",&N);

    //v[0]~v[N-1]に格納する
    double* v;
    NEW(v,N);
    for(int i=0;i<N;i++){
        scanf("%lf",&v[i]);
    }

    //マージソート用の仮配列
    double* u;
    NEW(u,N);
    //vをマージソート
    merge_sort(v,0,N-1,u);
    free(u);

    for(int i=0;i<=N-2;i++){
        v[i+1]=(v[i]+v[i+1])/2;
    }

    double ans=v[N-1];
    printf("%lf\n",ans);

    return 0;
}