You are given a string s, consisting of lowercase English letters, and the integer m.

One should choose some symbols from the given string so that any contiguous subsegment of length m has at least one selected symbol. Note that here we choose positions of symbols, not the symbols themselves.

Then one uses the chosen symbols to form a new string. All symbols from the chosen position should be used, but we are allowed to rearrange them in any order.

Formally, we choose a subsequence of indices 1 ≤ i1 < i2 < ... < it ≤ |s|. The selected sequence must meet the following condition: for every j such that 1 ≤ j ≤ |s| - m + 1, there must be at least one selected index that belongs to the segment [j, j + m - 1], i.e. there should exist a k from 1 to t, such that j ≤ ik ≤ j + m - 1.

Then we take any permutation p of the selected indices and form a new string sip1sip2... sipt.

Find the lexicographically smallest string, that can be obtained using this procedure.

Input

The first line of the input contains a single integer m (1 ≤ m ≤ 100 000).

The second line contains the string s consisting of lowercase English letters. It is guaranteed that this string is non-empty and its length doesn't exceed 100 000. It is also guaranteed that the number m doesn't exceed the length of the string s.

Output

Print the single line containing the lexicographically smallest string, that can be obtained using the procedure described above.

Examples

Input

3
cbabc


Output

a


Input

2
abcab


Output

aab


Input

3
bcabcbaccba


Output

aaabb

Note

In the first sample, one can choose the subsequence {3} and form a string "a".

In the second sample, one can choose the subsequence {1, 2, 4} (symbols on this positions are 'a', 'b' and 'a') and rearrange the chosen symbols to form a string "aab".
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5;
int mark[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int m;
  string s;
  cin >> m >> s;
  int n = s.size();
  string ans = "";
  int marked = m - 1;
  for (int i = 0; i < marked; ++i) {
    mark[i] = 1;
  }
  for (char c = 'a'; c <= 'z'; ++c) {
    int cnt = 0;
    int opt = -1;
    for (int i = 0; i < n; ++i) {
      if (s[i] == c) {
        ++cnt;
        opt = i;
      }
      if (mark[i] == 0 && opt >= 0) {
        ans.push_back(c);
        --cnt;
        for (int j = opt; j < opt + m && j < n; ++j) {
          marked += 1 - mark[j];
          mark[j] = 1;
        }
        opt = -1;
      }
    }
    if (marked < n) {
      for (int i = 0; i < cnt; ++i) {
        ans.push_back(c);
      }
    }
  }
  cout << ans << '\n';
  return 0;
}
