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)

import org.omg.CORBA.INTERNAL;
import sun.reflect.generics.tree.Tree;

import java.io.*;
import java.math.BigInteger;
import java.text.DecimalFormat;
import java.util.*;

public class Main {

    static InputReader sc = new InputReader(System.in);
    static PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) throws IOException {
        solver s = new solver();
        long t = 1;
        while (t > 0) {
            s.solve();
            t--;
        }
        out.close();
    }

   /* static class descend implements Comparator<pair1> {
        public int compare(pair1 o1, pair1 o2) {
            if (o1.pop != o2.pop)
                return (int) (o1.pop - o2.pop);
            else
                return o1.in - o2.in;
        }
    }*/

    static class InputReader {
        public BufferedReader br;
        public StringTokenizer token;

        public InputReader(InputStream stream) {
            br = new BufferedReader(new InputStreamReader(stream), 32768);
            token = null;
        }

        public String next() {
            while (token == null || !token.hasMoreTokens()) {
                try {
                    token = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return token.nextToken();
        }

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

        public String nextLine() throws IOException {
            return br.readLine();
        }

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

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

    static class card {
        long a;
        int cnt;
        int i;

        public card(long a, int cnt, int i) {
            this.a = a;
            this.cnt = cnt;
            this.i = i;
        }
    }

    static class ascend implements Comparator<pair> {
        public int compare(pair o1, pair o2) {
            if (o1.a != o2.a)
                return (int) (o1.a - o2.a);
            else
                return o1.b - o2.b;
        }
    }

    static class extra {

        static boolean v[] = new boolean[100001];
        static List<Integer> l = new ArrayList<>();
        static int t;

        static void shuffle(int a[]) {
            for (int i = 0; i < a.length; i++) {
                int t = (int) Math.random() * a.length;
                int x = a[t];
                a[t] = a[i];
                a[i] = x;
            }
        }

        static long gcd(long a, long b) {
            if (b == 0)
                return a;
            else
                return gcd(b, a % b);
        }

        static boolean valid(int i, int j, int r, int c, char ch[][]) {
            if (i >= 0 && i < r && j >= 0 && j < c && ch[i][j] != '*') {
                // System.out.println(i+" /// "+j);
                return true;
            } else {
                //  System.out.println(i+" //f "+j);
                return false;
            }
        }

        static void seive() {
            for (int i = 2; i < 100001; i++) {
                if (!v[i]) {
                    t++;
                    l.add(i);
                    for (int j = 2 * i; j < 100001; j += i)
                        v[j] = true;
                }
            }
        }

        static int binary(LinkedList<Integer> a, long val, int n) {
            int mid = 0, l = 0, r = n - 1, ans = 0;
            while (l <= r) {
                mid = (l + r) >> 1;
                if (a.get(mid) == val) {
                    r = mid - 1;
                    ans = mid;
                } else if (a.get(mid) > val)
                    r = mid - 1;
                else {
                    l = mid + 1;
                    ans = l;
                }
            }
            return (ans + 1);
        }

        static long fastexpo(long x, long y) {
            long res = 1;
            while (y > 0) {
                if ((y & 1) == 1) {
                    res *= x;
                }
                y = y >> 1;
                x = x * x;
            }
            return res;
        }

        static long lfastexpo(long x, long y, long p) {
            long res = 1;
            x = x % p;
            while (y > 0) {
                if ((y & 1) == 1) {
                    res = (res * x) % p;
                }
                y = y >> 1;
                x = (x * x) % p;
            }
            return res;
        }

       /* void dijsktra(int s, List<pair> l[], int n) {
            PriorityQueue<pair> pq = new PriorityQueue<>(new ascend());
            int dist[] = new int[100005];
            boolean v[] = new boolean[100005];
            for (int i = 1; i <= n; i++)
                dist[i] = 1000000000;
            dist[s] = 0;
            for (int i = 1; i < n; i++) {
                if (i == s)
                    pq.add(new pair(s, 0));
                else
                    pq.add(new pair(i, 1000000000));
            }
            while (!pq.isEmpty()) {
                pair node = pq.remove();
                v[node.a] = true;
                for (int i = 0; i < l[node.a].size(); i++) {
                    int v1 = l[node.a].get(i).a;
                    int w = l[node.a].get(i).b;
                    if (v[v1])
                        continue;
                    if ((dist[node.a] + w) < dist[v1]) {
                        dist[v1] = dist[node.a] + w;
                        pq.add(new pair(v1, dist[v1]));
                    }
                }
            }
        }*/
    }

    static class pair {
        int a;
        int b;

        public pair(int a, int i) {
            this.a = a;
            this.b = i;
        }
    }

    static class pair1 {
        pair p;
        int in;

        public pair1(pair a, int n) {
            this.p = a;
            this.in = n;
        }
    }


    static int inf = 5000013;

    static class solver {
        DecimalFormat df = new DecimalFormat("0.000000");
        extra e = new extra();
        int mod = 1000000007;
        int in[][] = new int[2][2];
        List<pair> l = new ArrayList<>();

        void solve() {
            int n = sc.nextInt();
            int a[] = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = sc.nextInt();
            e.shuffle(a);
            Arrays.sort(a);
            double prev = (a[0] + a[1]) / 2.0;
            for (int i = 2; i < n; i++)
                prev = (prev + a[i]) / 2.0;
            System.out.println(df.format(prev));
        }
    }
}