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 m;
string s;
char mn[100010], mx;
int get(int l, int r) {
  if (r - l + 1 < m) return 0;
  for (int i = l + m - 1; i >= l; i--) {
    if (s[i] == mx) return 1 + get(i + 1, r);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> m;
  cin >> s;
  multiset<char> chs;
  for (int i = 0; i < m; i++) {
    chs.insert(s[i]);
  }
  for (int i = 0; i + m - 1 < s.size(); i++) {
    mn[i] = *(chs.begin());
    chs.erase(chs.equal_range(s[i]).first);
    if (i + m < s.size()) chs.insert(s[i + m]);
  }
  for (int i = 0; i + m - 1 < s.size(); i++) {
    mx = max(mx, mn[i]);
  }
  vector<int> chosen;
  vector<char> word;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] < mx) {
      chosen.push_back(i);
      word.push_back(s[i]);
    }
  }
  int cnt = 0;
  if (chosen.size() == 0) {
    cnt += get(0, (int)s.size() - 1);
  } else {
    cnt += get(0, chosen[0] - 1);
    for (int i = 0; i < (int)chosen.size() - 1; i++) {
      cnt += get(chosen[i] + 1, chosen[i + 1] - 1);
    }
    cnt += get(chosen[chosen.size() - 1] + 1, s.size() - 1);
  }
  for (int i = 0; i < cnt; i++) {
    word.push_back(mx);
  }
  sort(word.begin(), word.end());
  for (int i = 0; i < word.size(); i++) {
    cout << word[i];
  }
  cout << endl;
  return 0;
}
