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;
char A[200005] = {};
int vis[200005] = {};
int main() {
  int a;
  while (cin >> a) {
    cin >> A;
    int n = strlen(A);
    char maxr = 0;
    for (int i = 0; i <= n - a; i++) {
      char now = A[i];
      int nex = i;
      for (int j = i; j < a + i; j++) {
        if (A[j] <= now) {
          nex = j;
          now = A[j];
        }
      }
      maxr = max(maxr, now);
      i = nex;
      vis[i] = 1;
    }
    vector<char> ans;
    for (int i = 0; A[i]; i++) {
      if (vis[i]) ans.push_back(A[i]);
    }
    for (int i = 0; A[i]; i++) {
      if (!vis[i] && A[i] < maxr) {
        ans.push_back(A[i]);
      }
    }
    sort(ans.begin(), ans.end());
    for (int i = 0; i < ans.size(); i++) {
      cout << ans[i];
    }
    cout << '\n';
  }
  return 0;
}
