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 m;
char s[120000];
int n;
int us[120000];
int main() {
  scanf("%d", &m);
  scanf(" %s", s);
  n = strlen(s);
  for (int i = 0; i < 26; ++i) {
    for (int j = 0; j < n; ++j)
      if (s[j] == 'a' + i) us[j] = 2;
    int pr = -1;
    int lst = -m - 100;
    int fl = 0;
    for (int j = 0; j < n; ++j) {
      if (us[j] == 1) {
        if (j - pr <= m) {
          pr = j;
          continue;
        }
        if (j - lst <= m) {
          us[lst] = 1;
          pr = j;
        } else {
          fl = 1;
          break;
        }
      } else if (us[j] == 2) {
        if (j - pr <= m) {
          lst = j;
        } else if (j - lst <= m) {
          us[lst] = 1;
          pr = lst;
          lst = j;
        } else {
          fl = 1;
          break;
        }
      }
    }
    if (!fl && pr + m < n) {
      if (lst + m >= n) {
        us[lst] = 1;
      } else {
        fl = 1;
      }
    }
    if (fl) {
      for (int j = 0; j < n; ++j)
        if (us[j]) us[j] = 1;
    } else {
      string ans;
      for (int i = 0; i < n; ++i)
        if (us[i] == 1) ans += s[i];
      sort(ans.begin(), ans.end());
      cout << ans << "\n";
      return 0;
    }
  }
  return 0;
}
