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 main() {
  int m;
  cin >> m;
  string s;
  cin >> s;
  set<pair<char, int> > ex;
  for (int i = 0; i < m; ++i) {
    ex.insert(make_pair(s[i], -i));
  }
  char mxch = 'a' - 1;
  int cnt = 0;
  for (int i = m - 1; i <= s.length();) {
    auto p = ex.begin();
    if (mxch < p->first) {
      mxch = p->first;
      cnt = 0;
    }
    if (mxch == p->first) {
      cnt++;
    }
    int j = -p->second + m;
    for (; i <= j && i <= s.length(); ++i) {
      ex.erase(make_pair(s[i - m], -(i - m)));
      ex.insert(make_pair(s[i], -(i)));
    }
  }
  string ans = "";
  for (auto p : s) {
    if (p < mxch) {
      ans += p;
    }
  }
  for (int i = 0; i < cnt; ++i) {
    ans += mxch;
  }
  sort(ans.begin(), ans.end());
  cout << ans;
  return 0;
}
