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() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.setf(ios::fixed);
  cout.precision(20);
  int m;
  cin >> m;
  string s;
  cin >> s;
  int n = s.size();
  vector<int> cnt(26);
  vector<int> have(n);
  for (int c = 0; c < 26; ++c) {
    for (int i = 0; i < n; ++i) {
      if (s[i] - 'a' == c) {
        have[i] = 1;
        ++cnt[c];
      }
    }
    bool fl = 1;
    int lst = -1;
    for (int i = 0; i < n; ++i) {
      if (have[i]) {
        fl &= (i - lst - 1 < m);
        lst = i;
      }
    }
    fl &= (n - lst - 1 < m);
    if (!fl) {
      continue;
    } else {
      int lst = -1;
      int nxt = 0;
      for (int i = 0; i < n; ++i) {
        nxt = max(nxt, i + 1);
        while (nxt < n && !have[nxt]) ++nxt;
        if (have[i] && s[i] - 'a' == c && nxt - lst - 1 < m) {
          --cnt[c];
        } else if (have[i])
          lst = i;
      }
      for (int i = 0; i < 26; ++i) {
        for (int j = 0; j < cnt[i]; ++j) {
          cout << char('a' + i);
        }
      }
      return 0;
    }
  }
}
