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;
string str, ans;
int n, k;
set<int> Pos[256], NotSolved;
void GoFor(int i) {
  ans += str[i];
  while (true) {
    auto it = NotSolved.lower_bound(i);
    if (it == NotSolved.end() || (*it) >= (i + k)) break;
    NotSolved.erase(it);
  }
}
bool SolveGreedy(int c) {
  while (NotSolved.size()) {
    auto it = NotSolved.begin();
    auto it2 = Pos[c].upper_bound(*it);
    if (it2 == Pos[c].begin()) return false;
    --it2;
    if (*it2 <= *it - k) return false;
    GoFor(*it2);
    Pos[c].erase(it2);
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> k;
  cin >> str;
  n = str.length();
  for (int i = 0; i < n; ++i) {
    Pos[str[i]].insert(i);
  }
  for (int i = k - 1; i < n; ++i) NotSolved.insert(i);
  for (int c = 'a'; c <= 'z'; ++c) {
    if (!SolveGreedy(c)) {
      while (Pos[c].size()) {
        auto it = Pos[c].begin();
        GoFor(*it);
        Pos[c].erase(it);
      }
    } else
      break;
  }
  cout << ans << endl;
  return 0;
}
