Problem: Polycarp is mad about coding, that is why he writes Sveta encoded messages. He calls the median letter in a word the letter which is in the middle of the word. If the word's length is even, the median letter is the left of the two middle letters. In the following examples, the median letter is highlighted: contest, info. If the word consists of single letter, then according to above definition this letter is the median letter. 

Polycarp encodes each word in the following way: he writes down the median letter of the word, then deletes it and repeats the process until there are no letters left. For example, he encodes the word volga as logva.

You are given an encoding s of some word, your task is to decode it. 

Input

The first line contains a positive integer n (1 ≤ n ≤ 2000) — the length of the encoded word.

The second line contains the string s of length n consisting of lowercase English letters — the encoding.

Output

Print the word that Polycarp encoded.

Examples

Input

5
logva


Output

volga


Input

2
no


Output

no


Input

4
abba


Output

baba

Note

In the first example Polycarp encoded the word volga. At first, he wrote down the letter l from the position 3, after that his word looked like voga. After that Polycarp wrote down the letter o from the position 2, his word became vga. Then Polycarp wrote down the letter g which was at the second position, the word became va. Then he wrote down the letter v, then the letter a. Thus, the encoding looked like logva.

In the second example Polycarp encoded the word no. He wrote down the letter n, the word became o, and he wrote down the letter o. Thus, in this example, the word and its encoding are the same.

In the third example Polycarp encoded the word baba. At first, he wrote down the letter a, which was at the position 2, after that the word looked like bba. Then he wrote down the letter b, which was at the position 2, his word looked like ba. After that he wrote down the letter b, which was at the position 1, the word looked like a, and he wrote down that letter a. Thus, the encoding is abba.

Solution (in JAVA):

Solution:


/*
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 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.StringTokenizer;

/**
 *
 * @author is2ac
 */
public class B_CF {

    public static void main(String[] args) {
        FastScanner56 fs = new FastScanner56();
        PrintWriter pw = new PrintWriter(System.out);
        //int t = fs.ni();
        int t = 1;
        //Boolean[][][] dp = new Boolean[101][101][10001];
        for (int tc = 0; tc < t; tc++) {
            long n = fs.ni();
            String s = fs.next();
            StringBuilder back = new StringBuilder();
            StringBuilder front = new StringBuilder();
            boolean right = false;
            if (s.length()%2==0) right = true;
            for (int i = 1; i < s.length(); i++) {
                if (right) {
                    back.append(s.charAt(i));
                } else {
                    front.append(s.charAt(i));
                }
                right = !right;
            }
            front.reverse();
            front.append(s.charAt(0));
            front.append(back);
            pw.println(front);
            
        }
        pw.close();
    }

    static long recur(long n, boolean c, boolean m) {
        long d = n % 10;
        if (c) {
            d--;
        }
        if (m) {
            d--;
        }
        if (d <= -1 && n == 0) {
            return (long) (-1e9);
        }
        m = false;
        if (d <= -1) {
            d = 10 + d;
            m = true;
        }
        if (n == 0) {
            return 0;
        }
        long res = recur(n / 10, false, m) + d;
        if (d != 9) {
            res = Math.max(res, recur(n / 10, true, m) + d + 10);
        }
        return res;
    }

    static boolean isp(int x) {
        int y = (int) Math.sqrt(x);
        if (Math.pow(y, 2) == x) {
            return true;
        }
        return false;
    }
}

class FastScanner56 {

    BufferedReader br;
    StringTokenizer st;

    public FastScanner56() {
        br = new BufferedReader(new InputStreamReader(System.in), 32768);
        st = null;
    }

    String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

    int ni() {
        return Integer.parseInt(next());
    }

    int[] intArray(int N) {
        int[] ret = new int[N];
        for (int i = 0; i < N; i++) {
            ret[i] = ni();
        }
        return ret;
    }

    long nl() {
        return Long.parseLong(next());
    }

    long[] longArray(int N) {
        long[] ret = new long[N];
        for (int i = 0; i < N; i++) {
            ret[i] = nl();
        }
        return ret;
    }

    double nd() {
        return Double.parseDouble(next());
    }

    String nextLine() {
        String str = "";
        try {
            str = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }
}