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 = 1e2 + 5;
const int MAXM = 1e2 + 5;
int m;
string s;
void solve(char c) {
  int last = -1, cnt = 0, i;
  for (i = 0; i < s.length(); i++) {
    if (i - last == m + 1) {
      for (int j = 0; j < cnt; j++) {
        cout << c;
      }
      for (int j = i; j < s.length(); j++) {
        if (s[j] == c) {
          cout << c;
        }
      }
      break;
    }
    if (s[i] <= c) {
      last = i;
      if (s[i] == c) {
        cnt++;
      }
    }
  }
  if (i == s.length() && i - 1 - last < m + 1) {
    last = -1;
    int lastc = -1;
    for (int j = 0; j < s.length(); j++) {
      if (s[j] < c) {
        last = j;
      }
      if (s[j] == c) {
        lastc = j;
      }
      if (j - last == m) {
        last = lastc;
        cout << c;
      }
    }
  } else {
    solve(c + 1);
  }
}
int main() {
  int n, x, y, l, r, min_v;
  cin >> m >> s;
  solve('a');
  cout << endl;
  return 0;
}
