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>
using namespace std;
int bin_search(vector<int> *V, int find, int left, int right) {
  while (right - left > 1) {
    int middle = (left + right) / 2;
    if ((*V)[middle] == find) return middle;
    if ((*V)[middle] > find) {
      right = middle;
    } else
      left = middle;
  }
  if (abs(find - (*V)[left]) == abs(find - (*V)[right]))
    return left;
  else if (abs(find - (*V)[left]) < abs(find - (*V)[right]))
    return left;
  else
    return right;
}
void count_linear_search(vector<int> *v, int find) {
  for (int i(0); i < v->size(); i++) {
    if ((*v)[i] == find) {
      cout << i + 1 << " ";
    }
  }
}
int main() {
  int n;
  cin >> n;
  string V;
  cin >> V;
  int current_index = 0;
  int tmp = 2;
  while (current_index < V.length()) {
    cout << V[current_index];
    current_index += tmp;
    tmp++;
  }
  return 0;
}