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 s[100003];
int bo[100003] = {0};
char ans[100003];
set<pair<char, int> > S;
int main() {
  int m;
  scanf("%d", &m);
  scanf("%s", s);
  int i;
  for (i = 0; i < m; i++) {
    S.insert(pair<char, int>(s[i], -i));
  }
  char mi = 0;
  char mis;
  int k;
  set<pair<char, int> >::iterator it = S.begin();
  mis = it->first;
  k = -(it->second);
  bo[k] = 1;
  mi = max(mi, mis);
  for (; s[i]; i++) {
    S.insert(pair<char, int>(s[i], -i));
    S.erase(pair<char, int>(s[i - m], -(i - m)));
    set<pair<char, int> >::iterator it = S.begin();
    if (k <= i - m) {
      mis = it->first;
      k = -(it->second);
      bo[k] = 1;
      mi = max(mi, mis);
    } else {
      if (it->first < mis) {
        mis = it->first;
        k = -(it->second);
        bo[k] = 1;
        mi = max(mi, mis);
      }
    }
  }
  int kk = 0;
  for (int i = 0; s[i]; i++) {
    if (bo[i] || s[i] < mi) {
      ans[kk++] = s[i];
    }
  }
  sort(ans, ans + kk);
  for (int i = 0; i < kk; i++) printf("%c", ans[i]);
  printf("\n");
  return 0;
}
