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 n, m;
string s;
vector<int> q[26];
vector<int> ans;
vector<int> pans;
vector<int> need;
vector<char> ss;
int check() {
  if (pans.size() == 0) return 0;
  int head = 0;
  for (int i = 1; i <= n - m; i++) {
    while ((pans[head] - i > m || pans[head] - i < 0) && head < pans.size() - 1)
      head++;
    if (pans[head] - i > m || pans[head] - i < 0) return 0;
  }
  return 1;
}
int main() {
  cin >> m;
  m--;
  cin >> s;
  n = s.size();
  for (int i = 0; i < n; i++) q[s[i] - 'a'].push_back(i + 1);
  for (int i = 0; i < 26; i++) {
    pans = ans;
    for (int j = 0; j < q[i].size(); j++) pans.push_back(q[i][j]);
    sort(pans.begin(), pans.end());
    if (check()) {
      int head = 0, h1 = 0;
      for (int j = 1; j <= n - m; j++) {
        if (ans.size() == 0) {
          if (need.size() != 0 && (need[need.size() - 1] - j <= m &&
                                   need[need.size() - 1] - j >= 0))
            continue;
          while ((q[i][h1] - j > m || q[i][h1] - j < 0) && h1 + 1 < q[i].size())
            h1++;
          while (h1 + 1 < q[i].size() &&
                 (q[i][h1 + 1] - j <= m && q[i][h1 + 1] - j >= 0))
            h1++;
          need.push_back(q[i][h1]);
          continue;
        }
        while (ans[head] < j && head < ans.size() - 1) head++;
        if (ans[head] - j > m || ans[head] - j < 0) {
          if (need.size() != 0 && (need[need.size() - 1] - j <= m &&
                                   need[need.size() - 1] - j >= 0))
            continue;
          while ((q[i][h1] - j > m || q[i][h1] - j < 0) && h1 + 1 < q[i].size())
            h1++;
          while (h1 + 1 < q[i].size() &&
                 (q[i][h1 + 1] - j <= m && q[i][h1 + 1] - j >= 0))
            h1++;
          need.push_back(q[i][h1]);
        }
      }
      for (int j = 0; j < need.size(); j++) ans.push_back(need[j]);
      sort(ans.begin(), ans.end());
      break;
    } else
      ans = pans;
  }
  for (int i = 0; i < ans.size(); i++) ss.push_back(s[ans[i] - 1]);
  sort(ss.begin(), ss.end());
  for (int i = 0; i < ss.size(); i++) putchar(ss[i]);
  return 0;
}
