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)

/* ABC138-C
   Takatan */

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

typedef struct tnode{
    double value;
    struct tnode *left;
    struct tnode *right;
} tnode;

tnode* createTNode(double value);
void insertTNode(tnode *new, tnode **first);
void checkTree(tnode *first, double *value);
int main(void){
    int number, i;
    double value, min;
    tnode *new, *tree = NULL;
    min = 1000;

    scanf( "%d" , &number);
    for(i=0; i < number; i++){
        scanf( "%lf" , &value);
        if(value<min){
            min = value; /* 最小値を保存 */
        }
        new = createTNode(value); /* 新しいノードを作成 */
        insertTNode(new, &tree); /* 二分探索木に挿入 */
    }
    value = min;
    checkTree(tree, &value); /* 残る1個の具材の価値をvaludに保存 */
    printf( "%f\n" , value);
    return 0;
}

tnode* createTNode(double value){ /* 新しいノードを作成する関数 */
    tnode *new = malloc(sizeof(tnode));

    if(new==NULL){
        exit(2);
    }

    new->value = value; /* 価値をコピー */
    new->left = NULL;
    new->right = NULL;
    return(new);
}

void insertTNode(tnode *new, tnode **first){

    if(*first==NULL){
        *first = new; /* 1つ上が先端の時 */
    }else if((new->value)<((*first)->value)){
        /* 新しい材料の価値の方が小さいとき、左を探索 */
        insertTNode(new, &((*first)->left)); /* 再帰的に実行 */
    }else{
        /* 新しい材料の価値の方が大きいまたは同じとき、右を探索 */
        insertTNode(new, &((*first)->right)); /* 再帰的に実行 */
    }
}

void checkTree(tnode *first, double *value){
    if(first->left!=NULL){ /* 左に子がいる時 */
        checkTree(first->left, value); /* 再帰的に実行 */
    }
    *value = (*value + first->value)/2; /* 価値の低いものから順に合成 */
    if(first->right!=NULL){ /* 右に子がいる時 */
        checkTree(first->right, value); /* 再帰的に実行 */
    }
}