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;
void _dbg(string) { cout << endl; }
template <class H, class... T>
void _dbg(string s, H h, T... t) {
  int l = s.find(',');
  cout << s.substr(0, l) << " = " << h << ", ";
  _dbg(s.substr(l + 1), t...);
}
template <class T, class U>
ostream& operator<<(ostream& o, const pair<T, U>& p) {
  o << "(" << p.first << "," << p.second << ")";
  return o;
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& v) {
  o << "[";
  for (T t : v) {
    o << t << ",";
  }
  o << "]";
  return o;
}
const int N = 100000;
int n, m;
bool used[N + 5];
char s[N + 5];
int hoge(char c) {
  int crnt = 0;
  int last = -1;
  int ret = 0;
  for (int i = (int)(0); i < (int)(n); i++) {
    crnt++;
    if (s[i] == c) last = i;
    if (used[i]) crnt = 0;
    if (crnt == m) {
      if (i - last >= m) return -1;
      crnt = i - last;
      used[last] = true;
      ret++;
    }
  }
  return ret;
}
int main() {
  cin >> m >> s;
  n = strlen(s);
  for (int i = (int)(0); i < (int)(26); i++) {
    char c = i + 'a';
    int r = hoge(c);
    if (r == -1) {
      r = 0;
      for (int j = (int)(0); j < (int)(n); j++)
        if (s[j] == c) {
          r++;
          used[j] = true;
        }
    } else {
      i = 100;
    }
    for (int j = (int)(0); j < (int)(r); j++) printf("%c", c);
  }
  cout << endl;
  return 0;
}
