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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

/*
String[]の初期化、整数の型がlongになっているか
関数（splitとかcontainsとか）内では"+"とかは正規表現となされるので"\\+"とする
'+'とかは問題ない
配列のコピーはarr1=arr.clone()
HashSet<>[]はおのおの初期化した？？？？？
負の数の割り算は絶対値が小さいほうに切り捨てられるex.-1/2=0ので、ある値>kみたいな式だとバグらせやすいので注意
ある値>=kとして、切り上げた値をとるべき
toLowerCase()はアルファベット以外に対して行ってもエラーにはならない
ArrayDequeではpopとpushはStackと同じ操作になる
//import static java.lang.System.*; はしばらく使わなそう(使うとしてもexit(0)のためだけ)←return;でよさそう
最後にout.flush()またはout.close()をするのを忘れずに！
Point2D p2d=new Point2D.Double(d1,d2);
Line2D l2d=new Line2D.Double(x1,y1,x2,y2);
Line2D l2d_0=new Line2D.Double(x11,y11,x22,y22);
if(l2d.intersectsLine(l2d_0))
next_Permutationを使うときはソート済み配列かどうか確認！！
参照型の比較するときはプリミティブ型にキャストするか.compare,.equalsを使う！
ex.map.get(0)=100,map.get(1)=100でもmap.get(0)==map.get(1)がfalseになりうる
*/
//入力終了→Ctrl+D
import java.io.*;
import java.awt.*;
import java.util.*;
import java.awt.geom.*;
import static java.lang.Math.*;
public class Main {
    public static void main(String[] $) {
        int n=sc.nextInt();
        PriorityQueue<Double> pq=new PriorityQueue<>();
        for (int i = 0; i < n; i++) {
            pq.add(sc.nextDouble());
        }
        while (pq.size()>1){
            double p1=pq.poll();
            double p2=pq.poll();
            pq.add((p1+p2)/2);
        }
        out.println(pq.poll());
        out.close();
    }

    static long gcd(long a, long b) {
        return b == 0 ? a : gcd(b, a % b);
    }

    //クラスカル法(0-indexed) n頂点m辺
    static HashSet<Integer> Kruskal(int[] u, int[] v, long[] cost, int n) {
        int m = u.length;
        UF.Init(n);
        PriorityQueue<Integer> S = new PriorityQueue<>((p1, p2) -> Long.compare(cost[p1], cost[p2]));
        for (int i = 0; i < m; i++) {
            S.add(i);
        }
        //最小全域木をなす辺
        HashSet<Integer> F = new HashSet<>();
        while (!S.isEmpty()) {
            int e = S.poll();
            if (!UF.same(u[e], v[e])) {
                F.add(e);
                UF.unite(u[e], v[e]);
            }
        }
        return F;
    }

    //ワーシャルフロイド(0-indexed)
    static void WF(double[][] d,int kappa,int lambda) {
        int n = kappa + lambda;
        for (int k = 0; k < n; k++) {
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    if (d[i][k] != -1 && d[k][j] != -1) {
                        d[i][j] = d[i][j] > d[i][k] + d[k][j] ? d[i][k] + d[k][j] : d[i][j];
                    }
                }
            }
        }
    }

    //Union Find木(0-indexed)
    static class UF {
        static int[] d;

        static void Init(int size) {
            d = new int[size];
            for (int i = 0; i < size; i++) {
                d[i] = -1;
            }
        }

        static boolean unite(int x, int y) {
            x = find(x);
            y = find(y);
            if (x == y) return false;
            if (d[x] > d[y]) {
                int t = d[y];
                d[y] = d[x];
                d[x] = t;
            }
            d[x] += d[y];
            d[y] = x;
            return true;
        }

        static boolean same(int x, int y) {
            return find(x) == find(y);
        }

        static int find(int k) {
            return d[k] < 0 ? k : (d[k] = find(d[k]));
        }

        static int size(int k) {
            return -d[find(k)];
        }
    }

    static PrintWriter out = new PrintWriter(System.out);

    static class sc {
        static Scanner scanner = new Scanner(System.in);

        static String next() {
            return scanner.next();
        }

        static String nextLine() {
            return scanner.nextLine();
        }

        static int nextInt() {
            return Integer.parseInt(scanner.next());
        }

        static long nextLong() {
            return Long.parseLong(scanner.next());
        }

        static double nextDouble() {
            return Double.parseDouble(scanner.next());
        }

        static boolean hasNext() {
            return scanner.hasNext();
        }

        static boolean hasNextInt() {
            return scanner.hasNextInt();
        }

        static boolean hasNext(String str) {
            return scanner.hasNext(str);
        }
    }

}