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;
int m;
string s, sol;
bool u[100005];
bool verif(char ch, int tip) {
  int ui = -1, uch = -1;
  for (int j = 0; j < s.size(); j++) {
    if (u[j] == 1) {
      ui = j;
    }
    if (s[j] == ch) {
      uch = j;
      if (tip == 2) {
        u[j] = 1;
      }
    }
    if (j - ui >= m) {
      ui = uch;
      if (tip == 1) {
        u[ui] = 1;
      }
    }
    if (j - ui >= m) {
      if (tip == 0) {
        return false;
      }
    }
  }
  return true;
}
int main() {
  cin >> m;
  cin >> s;
  for (char i = 'a'; i <= 'z'; i++) {
    if (verif(i, 0)) {
      verif(i, 1);
    } else {
      verif(i, 2);
    }
  }
  for (int j = 0; j < s.size(); j++) {
    if (u[j] == 1) {
      sol += s[j];
    }
  }
  sort(sol.begin(), sol.end());
  cout << sol;
  return 0;
}
