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;
template <typename T>
T& input(T& v) {
  for (int i = 0; i < v.size(); i++) cin >> v[i];
  return v;
}
template <typename T>
T& input_2d(T& v) {
  int N = v.size();
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < v[i].size(); j++) cin >> v[i][j];
  }
}
template <typename T>
T& input_pair(T& v) {
  for (int i = 0; i < v.size(); i++) cin >> v[i].first >> v[i].second;
  return v;
}
template <typename T>
void print(const T& v) {
  int N = v.size();
  for (int i = 0; i < N; i++) cout << v[i] << " ";
  cout << '\n';
}
template <typename T>
void print_2d(T& v) {
  int N = v.size();
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < v[i].size(); j++) cout << v[i][j] << " ";
    cout << '\n';
  }
}
template <typename T>
void print_pair(const T& v) {
  int N = v.size();
  for (int i = 0; i < N; i++) cout << v[i].first << " " << v[i].second << '\n';
}
template <typename T>
inline int digits(const T& N) {
  return (N > 0 ? (int)((log10(N)) + 1) : 1);
}
inline bool ispoweroftwo(int x) { return (x && (!(x & (x - 1)))); }
bool if_integer(double x) {
  if (x - floor(x) == 0 && x > 0)
    return true;
  else
    return false;
}
int binomial(int n, int k) {
  vector<int> C(k + 1, 0);
  C[0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = min(i, k); j > 0; j--) C[j] = C[j] + C[j - 1];
  }
  return C[k];
}
inline double distance(int a, int b, int c, int d) {
  return sqrt(((long long)(a - c)) * (a - c) + ((long long)(b - d)) * (b - d));
}
class compare {
  bool operator()(pair<int, int>& p1, pair<int, int>& p2) {
    if (p1.first == p2.first) return p1.second > p2.second;
    return p1.first > p2.first;
  }
};
bool sortbysec(const pair<int, int>& a, const pair<int, int>& b) {
  if (a.first == b.first) return (a.second < b.second);
  return (a.first > b.first);
}
int BS(vector<int>& arr, int x) {
  int l = 0, r = arr.size() - 1;
  while (l <= r) {
    int m = l + (r - l) / 2;
    if (arr[m] == x) return m;
    if (arr[m] < x)
      l = m + 1;
    else
      r = m - 1;
  }
  return -1;
}
bool mysort(const pair<int, int>& a, const pair<int, int>& b) {
  if (a.first == b.first) return (a.second < b.second);
  return (a.first > b.first);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  string s, second;
  cin >> s;
  int l = s.size(), count = 0;
  if (l % 2 == 0) {
    int i = l - 1;
    while (count < l) {
      second.push_back(s[i]);
      if (count % 2 == 0)
        i = l - 1 - i;
      else
        i = l - 1 - i - 1;
      count++;
    }
    reverse(second.begin(), second.end());
    cout << second;
  } else {
    int i = 0;
    while (count < l) {
      second.push_back(s[i]);
      if (count % 2 == 0)
        i = l - 1 - i;
      else
        i = l - 1 - i + 1;
      count++;
    }
    reverse(second.begin(), second.end());
    cout << second;
  }
}