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;
deque<int> de;
char s[100005];
char ans[100005];
bool used[100005];
int main() {
  int n, i, k;
  scanf("%d", &k);
  scanf("%s", s + 1);
  n = strlen(s + 1);
  int last = 0;
  int cnt = 0;
  char mx = 'a';
  for (i = 1; i <= n; i++) {
    while (de.empty() == false && s[de.back()] >= s[i] && s[i] >= mx) {
      de.pop_back();
    }
    if (s[i] < mx) {
      ans[++cnt] = s[i];
      used[i] = 1;
      last = i;
    } else {
      de.push_back(i);
    }
    while (de.empty() == false && i - de.front() >= k) {
      de.pop_front();
    }
    if (i >= k && i - last == k) {
      ans[++cnt] = s[de.front()];
      used[de.front()] = 1;
      mx = max(mx, s[de.front()]);
      last = de.front();
    }
  }
  for (i = 1; i <= n; i++) {
    if (s[i] < mx && used[i] == 0) {
      ans[++cnt] = s[i];
    }
  }
  sort(ans + 1, ans + cnt + 1);
  printf("%s", ans + 1);
  return 0;
}
