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;
char ch[110000];
int m, pd[110000], n, pre[110000];
int check(int K) {
  memcpy(pre, pd, sizeof pd);
  int now = 1;
  while (now <= n && pd[now]) now++;
  for (int i = 1; i <= n; i++)
    if (ch[i] == K) {
      now = max(now, i - m + 1);
      while (now <= n - m + 1 && now <= i) pd[now] = 1, now++;
    }
  for (int i = 1; i <= n - m + 1; i++)
    if (pd[i] == 0) return 0;
  memcpy(pd, pre, sizeof pre);
  return 1;
}
void getans(int K) {
  for (int i = 'a'; i < K; i++)
    for (int j = 1; j <= n; j++)
      if (ch[j] == i) putchar(i);
  int now = 1;
  while (now <= n && pd[now]) now++;
  int pre = 0;
  for (int i = 1; i <= n; i++) {
    if (ch[i] == K) {
      if (i - m + 1 > now) {
        while (now <= n - m + 1 && now <= pre) pd[now] = 1, now++;
        while (now <= n - m + 1 && pd[now]) now++;
        putchar(K);
      }
      pre = i;
    }
  }
  if (now <= n - m + 1) putchar(K);
}
int main() {
  scanf("%d", &m);
  scanf("%s", ch + 1);
  n = strlen(ch + 1);
  for (int i = 'a'; i <= 'z'; i++)
    if (check(i)) {
      getans(i);
      return 0;
    }
}
