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 ans['z' + 2];
int T['z' + 2];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int m;
  string s;
  cin >> m >> s;
  int n = s.size();
  set<pair<char, int> > se;
  int d = -1;
  for (int r = 0, l = 0; r < n; ++r) {
    ++T[int(s[r])];
    while (l <= d) {
      se.erase(pair<char, int>(s[l], -l));
      ++l;
    }
    se.insert(pair<char, int>(s[r], -r));
    if (r - l + 1 < m) continue;
    auto x = se.begin();
    ++ans[int(x->first)];
    d = -x->second;
  }
  d = 0;
  for (int i = 'z'; i >= 'a'; --i) {
    if (d || ans[i + 1]) ans[i] = T[i], d = 1;
  }
  for (int i = 'a'; i <= 'z'; ++i) {
    for (int j = 0; j < ans[i]; ++j) cout << char(i);
  }
  cout << endl;
  return 0;
}
