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;
int c[31];
int a[100111][31];
int main() {
  int m;
  string s;
  cin >> m;
  cin >> s;
  s = '0' + s;
  int n = s.length() - 1;
  for (int j = 0; j <= 26; j++) a[0][j] = -1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= 26; j++) a[i][j] = a[i - 1][j];
    a[i][s[i] - 'a' + 1] = i;
    c[s[i] - 'a']++;
  }
  for (int p = 0; p < 26; p++) {
    int cnt = 0;
    int last = -1;
    bool ok = true;
    for (int i = m; i <= n; i++) {
      ok = false;
      for (int j = 0; j <= p; j++) {
        if (a[i][j] > i - m) {
          ok = true;
          break;
        }
      }
      if (!ok) {
        if (last > i - m)
          ok = true;
        else {
          if (a[i][p + 1] > i - m) {
            last = a[i][p + 1];
            ok = true;
            cnt++;
          } else
            break;
        }
      }
    }
    if (ok) {
      for (int i = 0; i < p; i++)
        for (int j = 0; j < c[i]; j++) cout << char('a' + i);
      for (int i = 0; i < cnt; i++) cout << char('a' + p);
      return 0;
    }
  }
  return 0;
}
