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.*;
import java.io.*;
public class rcipher
{
	int n ;
	String s ;
	StringBuffer s1 ;
	Scanner sc=new Scanner(System.in);
	PrintWriter pr=new PrintWriter(System.out,true);
	public static void main(String... args)
	{
       rcipher c=new rcipher();
       c.prop();
	}
	public void prop()
	{
		int total ;
		n=sc.nextInt();
		s=sc.next();
         total=0 ;
         s1=new StringBuffer("");
         s1.append(s.charAt(0));
         
		for (int i=1;total<n-1 ;++i) {
               
			 s1.append(s.charAt(total+1));
             total=total+1*(i+1) ;
              
		}
        pr.println(s1);
	}
}