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 n, m;
string s;
bool puede(int c) {
  char car = 'a' + c;
  int last = -1;
  for (int i = 0; i < n; i++)
    if (s[i] <= car)
      last = i;
    else if (i - last >= m)
      return false;
  return true;
}
int sol[26];
void escribe(int c) {
  char car = 'a' + c;
  int last = -1;
  int lastc = -1;
  for (int i = 0; i < n; i++) {
    if (s[i] < car) {
      sol[s[i] - 'a']++;
      last = i;
    } else {
      if (s[i] == car) lastc = i;
      if (i - last >= m) {
        sol[c]++;
        last = lastc;
      }
    }
  }
  for (int i = 0; i < 26; i++) cout << string(sol[i], char('a' + i));
  cout << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> m >> s;
  n = int(s.size());
  for (int c = 0; c < 26; c++) {
    if (puede(c)) {
      escribe(c);
      exit(0);
    }
  }
}
