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 main() {
  int m;
  string s;
  cin >> m >> s;
  string answer = "";
  int n = s.length();
  vector<char> filled(n, false);
  for (char c = 'a'; c <= 'z'; c++) {
    vector<char> old = filled;
    int nc = 0;
    vector<int> pos;
    vector<char> filledNow(n, false);
    for (int i = 0; i < n; i++)
      if (s[i] == c) {
        pos.push_back(i);
        nc++;
        int j = i;
        while (j >= 0 && j > i - m && !filledNow[j]) {
          filled[j] = filledNow[j] = true;
          j--;
        }
      }
    bool fail = false;
    for (int i = 0; i < n - m + 1; i++)
      if (!filled[i]) {
        fail = true;
        break;
      }
    if (fail) {
      for (int i = 0; i < nc; i++) answer += c;
      continue;
    }
    filled = old;
    int minEmpty = 0;
    while (minEmpty < n - m + 1 && filled[minEmpty]) minEmpty++;
    int nAdd = 0;
    filledNow.assign(n, false);
    for (int i = 0; i < nc; i++) {
      if (minEmpty >= n - m + 1) break;
      if (i + 1 < nc && pos[i + 1] - m < minEmpty) continue;
      nAdd++;
      int j = pos[i];
      while (j >= 0 && j > pos[i] - m && !filledNow[j]) {
        filled[j] = filledNow[j] = true;
        j--;
      }
      while (minEmpty < n && filled[minEmpty]) minEmpty++;
      if (minEmpty >= n - m + 1) break;
    }
    for (int i = 0; i < nAdd; i++) answer += c;
    break;
  }
  cout << answer << endl;
  return 0;
}
