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;
void solve() {
  int m;
  string s;
  cin >> m;
  cin >> s;
  cin >> s;
  char c = 'a';
  set<int> pos;
  string ans;
  bool ok = false;
  while (!ok) {
    int i = 0;
    set<int> curPos;
    for (int i = 0; i < s.size(); ++i)
      if (s[i] == c) curPos.insert(i);
    while (i <= s.size() - m) {
      bool curTaken = false;
      auto it = pos.end();
      if (pos.size()) {
        it = pos.upper_bound(i + m - 1);
        if (it == pos.begin())
          it = pos.end();
        else
          --it;
      }
      if (it == pos.end() || *it < i) {
        if (curPos.size()) {
          it = curPos.upper_bound(i + m - 1);
          if (it != curPos.begin()) {
            --it;
            curTaken = true;
          } else
            it = pos.end();
        }
      }
      if (it == pos.end() || *it < i) {
        for (int i = 0; i < curPos.size(); ++i) ans += c;
        ++c;
        pos.insert(curPos.begin(), curPos.end());
        break;
      } else {
        if (curTaken) {
          ans += c;
          pos.insert(*it);
          curPos.erase(it);
        }
        i = *it + 1;
      }
    }
    if (i > s.size() - m) {
      ok = true;
    }
  }
  cout << ans << endl;
}
int main() {
  solve();
  return 0;
}
