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;
unordered_map<char, int> seq;
bool vis[(int)1e5 + 5], A[(int)1e5 + 5];
int main() {
  int m;
  string s;
  cin >> m >> s;
  char c;
  for (c = 'a'; c <= 'z'; c++) {
    seq.clear();
    memcpy(A, vis, sizeof(A));
    for (int j = 0; j < m; j++) seq[s[j]]++;
    if (seq[c]) vis[0] = true;
    for (int j = 1; j + m - 1 < s.size(); j++) {
      seq[s[j - 1]]--;
      seq[s[j + m - 1]]++;
      if (seq[c]) vis[j] = true;
    }
    if (count(vis, vis + (int)s.size(), true) == s.size() - m + 1) break;
  }
  for (char t = 'a'; t < c; t++) {
    int v = count(s.begin(), s.end(), t);
    while (v--) cout << t;
  }
  for (int i = 0; i <= s.size() - m; i++) {
    if (A[i]) continue;
    int t = i;
    for (int j = i; j < i + m && j < s.size(); j++)
      if (s[j] == c) t = j;
    cout << c;
    i = t;
  }
  cout << endl;
  return 0;
}
