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>
#pragma gcc optimize("Ofast,no-stack-protector,tune=native")
#pragma gcc optimize("sse,sse2,sse3,sse4,ssse3")
#pragma gcc optimize("O3")
#pragma gcc optimize("abm,mmx,avx,avx2,unroll-loops,fast-math,section-anchors")
using namespace std;
vector<long long> solve(string s, long long m, char c) {
  vector<long long> marks;
  long long l = 0, r = 0, id = -1;
  while (r < s.size()) {
    while (r < s.size() && r - l < m) {
      if (s[r] < c) {
        marks.push_back(r);
        l = r + 1;
      } else if (s[r] == c)
        id = r;
      r++;
    }
    if (r - l == m) {
      if (id < l) return solve(s, m, ++c);
      if (id - l <= m) {
        marks.push_back(id);
        l = id + 1;
      } else
        return solve(s, m, ++c);
    }
  }
  return marks;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long m;
  cin >> m;
  string s;
  cin >> s;
  auto tmp = solve(s, m, 'a');
  multiset<char> ans;
  for (auto it : tmp) ans.insert(s[it]);
  for (auto it : ans) cout << it;
}
