Problem: Polycarp loves ciphers. He has invented his own cipher called repeating.

Repeating cipher is used for strings. To encrypt the string s=s_{1}s_{2} ... s_{m} (1 ≤ m ≤ 10), Polycarp uses the following algorithm:

  * he writes down s_1 ones, 
  * he writes down s_2 twice, 
  * he writes down s_3 three times, 
  * ... 
  * he writes down s_m m times. 



For example, if s="bab" the process is: "b" → "baa" → "baabbb". So the encrypted s="bab" is "baabbb".

Given string t — the result of encryption of some string s. Your task is to decrypt it, i. e. find the string s.

Input

The first line contains integer n (1 ≤ n ≤ 55) — the length of the encrypted string. The second line of the input contains t — the result of encryption of some string s. It contains only lowercase Latin letters. The length of t is exactly n.

It is guaranteed that the answer to the test exists.

Output

Print such string s that after encryption it equals t.

Examples

Input


6
baabbb


Output


bab

Input


10
ooopppssss


Output


oops

Input


1
z


Output


z

Solution (in JAVA):

Solution:


import java.lang.*;
import java.util.*;

public class Main {

    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        StringBuilder t = new StringBuilder(sc.nextLine());
        int n = t.length();
        StringBuilder sb = new StringBuilder("");
        while (n!= 0) {
            if (n % 2 == 0) {
                sb.append(t.charAt(t.length() - 1));
                t.deleteCharAt(t.length() - 1);
            } else {
                sb.append(t.charAt(0));
                t.deleteCharAt(0);
            }
            n--;

        }
        System.out.println(sb.reverse());

    }
}