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>
const int MAX_N = (int)1e5 + 123;
using namespace std;
int m;
string second;
vector<int> id[30];
set<int> q;
bool check(set<int> &q) {
  int last = -1;
  for (auto i : q) {
    if (i - last > m) return 0;
    last = i;
  }
  if (int((second).size()) - last > m) return 0;
  return 1;
}
void out(set<int> &q) {
  vector<char> ans;
  for (auto i : q) ans.push_back(second[i]);
  sort(ans.begin(), ans.end());
  for (auto i : ans) cout << i;
  exit(0);
}
int main() {
  ios_base ::sync_with_stdio(NULL), cin.tie(NULL), cout.tie(NULL);
  cin >> m >> second;
  for (int i = 0; i < int((second).size()); i++)
    id[second[i] - 'a'].push_back(i);
  for (int i = 0; i < 26; i++) {
    for (auto j : id[i]) q.insert(j);
    if (check(q)) {
      for (auto j : id[i]) q.erase(j);
      int now = 0, j = 0;
      while (now + m - 1 < int((second).size())) {
        auto it = q.lower_bound(now + m);
        if (it != q.begin()) {
          it--;
          if (*it >= now) {
            now = *it + 1;
            continue;
          }
        }
        while (j + 1 < int((id[i]).size()) && id[i][j + 1] < now + m) j++;
        assert(id[i][j] >= now);
        q.insert(id[i][j]);
        now = id[i][j++] + 1;
      }
      out(q);
    }
  }
  return 0;
}
