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 CPP):

Solution:
#include <bits/stdc++.h>
char s[55], c[55];
int main() {
  while (~scanf("%s", s)) {
    int l = strlen(s);
    if (l == 1) {
      printf("%s\n", s);
      continue;
    }
    if (l % 2 == 0) {
      int i = 0, j = l - 1, k = l - 1;
      while (i < j) {
        c[k--] = s[j];
        c[k--] = s[i];
        i++;
        j--;
      }
    } else {
      int i = 0, j = l - 1, k = l - 1;
      while (j + i <= l) {
        c[k--] = s[i];
        if (k < 0) break;
        c[k--] = s[j];
        if (k < 0) break;
        i++;
        j--;
      }
    }
    for (int i = 0; i < l; i++) {
      printf("%c", c[i]);
    }
    printf("\n");
    memset(s, 0, sizeof(s));
    memset(c, 0, sizeof(c));
  }
  return 0;
}