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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class EjercicioH1 {

	public static void main(String[] args) 
	{
		// TODO Auto-generated method stub
		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
		try
		{
			String palabra = bf.readLine();
			int mitad = palabra.length()/2;
			if(palabra.length()%2 == 0)
			{
				mitad--;
			}
			char[]y = palabra.toCharArray();
			char[]x = new char[palabra.length()];
			x[0] = y[mitad];
			int cont =1;
			int i = mitad +1;
			int j = mitad-1;
			
			while(i<y.length || j>=0)
			{
				if(cont%2 == 1)
				{
					x[cont] = y[i];
					i++;
					cont++;
				}
				else
				{
					x[cont] = y[j];
					j--;
					cont++;
				}
			}

			

			String resp = new String(x);
			System.out.println(resp);
			bf.close();
		}
		catch(IOException e)
		{
			
		}
	}

}