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;
bool calc(int n, const string &s, int p) {
  int last = -1, l = -1, sum = 0, len = s.length();
  bool ised = 1;
  for (int i = 0; i < len; ++i) {
    if (s[i] <= p) {
      last = i;
      if (s[i] == p) ++sum;
    }
    if (i - last >= n) ised = 0;
  }
  if (!ised) return cout << string(sum, p), 0;
  last = -1, sum = 0;
  for (int i = 0; i < len; ++i) {
    if (s[i] < p)
      l = i, last = i;
    else if (s[i] == p)
      l = i;
    if (i - last >= n) last = l, ++sum;
  }
  cout << string(sum, p);
  return 1;
}
int main() {
  int n;
  string s;
  cin >> n >> s;
  for (int i = 'a'; i <= 'z' && !calc(n, s, i); ++i)
    ;
}
