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>
int nxt[100005];
char s[100005];
int main() {
  int n, m;
  scanf("%d\n%s", &m, s + 1);
  n = strlen(s + 1);
  int num, i;
  for (i = 0; i < 26; i++) {
    int last = n + 1, j;
    for (j = n; j >= 1; j--) {
      nxt[j] = last;
      if (s[j] - 'a' <= i) last = j;
    }
    last = num = 0;
    for (j = 1; j <= n; j++)
      if (s[j] - 'a' > i && j - last >= m)
        break;
      else if (s[j] - 'a' == i && nxt[j] - last <= m)
        continue;
      else if (s[j] - 'a' <= i)
        last = j, num += (s[j] - 'a') == i;
    if (j > n) break;
    for (j = 1; j <= n; j++)
      if (s[j] - 'a' == i) printf("%c", s[j]);
  }
  for (int j = 1; j <= num; j++) printf("%c", 'a' + i);
}
