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.util.*;

public class Main {

    public static void main(String[] args) {
        Scanner Input = new Scanner(System.in);
        int N = Input.nextInt();
        String Po = Input.next();
        ArrayList<Integer> A = new ArrayList();
        int Sum = 0;
        for (int I = 1; I <= N; I++) {
            Sum += I;
            A.add(Sum);
        }
        //System.out.println(A);
        String Ans = "";
        try {
            for (int I = 0; I < N; I++) {
                Ans += Po.charAt(A.get(I)-1);
            }
        }catch(Exception e){}
        System.out.println(Ans);
    }
}