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.Scanner;

public class solve {

    public static void main(String[] args) {
        int l, i, j;
        char[] a = new char[50];
        Scanner in = new Scanner(System.in);
        String s = in.nextLine();
        a = s.toCharArray();
        l = s.length();
        char[] x = new char[55];
        if (l % 2 == 0) {
            for (i = 0, j = 0; i <= (l - 1) / 2 && j < l; i++, j++) {
                x[j] = a[l - 1 - i];
                j = j + 1;
                x[j] = a[i];
            }
        } else {
            for (i = 0, j = 0; i <= (l - 1) / 2 && j < l; i++, j++)
                {
                    x[j] = a[i];
                    j = j + 1;
                    x[j] = a[l - 1 - i];
                }
            }
        for(i=l-1;i>=0;i--){
            System.out.printf("%c",x[i]);
        }
        }
    }