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;
const int MAXN = 100000;
vector<int> pos[26];
char s[MAXN + 10];
int l[MAXN + 10], r[MAXN + 10];
int main() {
  int n, m;
  scanf("%d%s", &m, s + 1);
  n = strlen(s + 1);
  for (int i = 1; i <= n; i++) l[i] = i - 1, r[i] = i + 1;
  r[0] = 1;
  for (int i = 1; i <= n; i++) pos[s[i] - 'a'].push_back(i);
  for (int i = 25; i >= 0; i--) {
    int sz = pos[i].size();
    int cnt = 0;
    for (int j = 0; j < sz; j++) {
      int del_len = r[pos[i][j]] - l[pos[i][j]];
      if (del_len > m) continue;
      r[l[pos[i][j]]] = r[pos[i][j]];
      l[r[pos[i][j]]] = l[pos[i][j]];
      cnt++;
    }
    if (cnt != sz) break;
  }
  int u = 0;
  vector<char> vec;
  while (u <= n) {
    if (u != 0) vec.push_back(s[u]);
    u = r[u];
  }
  sort(vec.begin(), vec.end());
  int vsz = vec.size();
  for (int i = 0; i < vsz; i++) printf("%c", vec[i]);
  printf("\n");
  return 0;
}
