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 len, lim;
char str[100005];
int pos[100005], posn, cnt[255];
bool vis[100005];
int main() {
  scanf("%d", &lim);
  scanf("%s", &str);
  len = strlen(str);
  for (int i = 0; i < len; i++) {
    cnt[str[i]]++;
  }
  for (int i = 'a'; i <= 'z'; i++) {
    memset(vis, 0, sizeof(vis));
    posn = 0;
    for (int j = 0; j < len; j++) {
      if (str[j] <= i) {
        pos[posn++] = j;
      }
    }
    int left = 0;
    for (int j = 0; j < posn; j++) {
      for (int k = max(left, pos[j] - lim + 1); k <= min(len, pos[j]); k++) {
        vis[k] = true;
      }
      left = pos[j];
    }
    for (int k = 0; k < len - lim + 1; k++) {
      if (!vis[k]) {
        goto end;
      }
    }
    {
      memset(vis, 0, sizeof(vis));
      posn = 0;
      for (int j = 0; j < len; j++) {
        if (str[j] < i) {
          pos[posn++] = j;
        }
      }
      int left = 0;
      for (int j = 0; j < posn; j++) {
        for (int k = max(left, pos[j] - lim + 1); k <= min(len, pos[j]); k++) {
          vis[k] = true;
        }
        left = pos[j];
      }
      int count = 0;
      for (int k = 0; k < len - lim + 1; k++) {
        if (!vis[k]) {
          for (int j = min(k + lim - 1, len - 1); j >= k; j--) {
            if (str[j] == i) {
              count++;
              k = j;
              break;
            }
          }
        }
      }
      for (int j = 'a'; j < i; j++) {
        for (int k = 0; k < cnt[j]; k++) {
          putchar(j);
        }
      }
      while (count--) {
        putchar(i);
      }
      return 0;
    }
  end:;
  }
  return 0;
}
