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() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  bool fin = false;
  int n, l, cnt[26] = {}, lim = 0;
  string s;
  vector<pair<char, int>> a, ans, tmp;
  cin >> n >> s;
  l = s.length();
  for (int(i) = 0; (i) < (l); ++(i)) {
    a.push_back(make_pair(s.at(i), i));
  }
  for (int(i) = 0; (i) < (n - 1); ++(i)) {
    cnt[s.at(i) - 'a']++;
  }
  for (int(i) = (n - 1); (i) < (l); ++(i)) {
    cnt[s.at(i) - 'a']++;
    int tm;
    for (int(j) = 0; (j) < (26); ++(j)) {
      if (cnt[j] > 0) {
        tm = j;
        break;
      }
    }
    cnt[s.at(i + 1 - n) - 'a']--;
    lim = max(lim, tm);
  }
  ans.push_back(make_pair('a', -1));
  for (int(i) = 0; (i) < (l); ++(i)) {
    if (a[i].first - 'a' <= lim) ans.push_back(a[i]);
  }
  ans.push_back(make_pair('a', l));
  auto it = ans.begin() + 1;
  while (it < ans.end() - 1) {
    while (it < ans.end() - 1 and (it->first) - 'a' == lim and
           ((it + 1)->second - (it - 1)->second) <= n) {
      ans.erase(it);
    }
    it++;
  }
  ans.erase(ans.begin());
  ans.erase(ans.end() - 1);
  sort(ans.begin(), ans.end());
  for (int(i) = 0; (i) < (ans.size()); ++(i)) {
    cout << ans[i].first;
  }
  cout << endl;
  return 0;
}
