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;
string inttostr(long long a) {
  string ans = "";
  if (a == 0) return "0";
  stringstream ss;
  while (a > 0) {
    string ch;
    ss << a % 10;
    ss >> ch;
    ss.clear();
    ans = ch + ans;
    a /= 10;
  }
  return ans;
}
long long strtoint(string s) {
  long long ans = 0;
  for (long long i = s.size() - 1; i >= 0; --i) {
    long long o = s[i] - '0';
    if (s[i] == 0) o = 0;
    ans += o * pow(10, s.size() - 1 - i);
  }
  return ans;
}
signed main() {
  string s;
  cin >> s;
  string ans = "";
  long long L = 0, R = s.size() - 1;
  if (s.size() % 2 == 1) {
    ans += s[L];
    L++;
  }
  while (L < R) {
    ans = s[R] + ans;
    ans = s[L] + ans;
    L++;
    R--;
  }
  cout << ans;
}