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;
const int DIM = 1e5 + 5;
const int SIGMA = 26;
vector<int> lst[SIGMA];
int mrk[DIM];
int main(void) {
  ios::sync_with_stdio(false);
  int m;
  string str, ans;
  cin >> m >> str;
  str = " " + str;
  for (int i = 1; i < str.length(); i++) lst[str[i] - 'a'].push_back(i);
  for (int k = 0; k < SIGMA; k++) {
    int mx = 0;
    for (int i = 1, j = -1; i < str.length(); i++) {
      if (mrk[i] == 1) mx = i;
      if (i - mx >= m) {
        while (j < int(lst[k].size()) - 1 && lst[k][j + 1] <= i) j++;
        if (j == -1) {
          for (int x : lst[k]) mrk[x] = 1;
          break;
        }
        if (i - lst[k][j] >= m) {
          for (int x : lst[k]) mrk[x] = 1;
          break;
        } else {
          mrk[lst[k][j]] = 1;
          mx = lst[k][j];
        }
      }
    }
  }
  for (int i = 1; i < str.length(); i++) {
    if (mrk[i] == 1) ans.push_back(str[i]);
  }
  sort(ans.begin(), ans.end());
  cout << ans << endl;
  return 0;
}
