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;
int n, m;
char s[100005];
vector<int> v[26];
int idx[26];
vector<int> w;
int main() {
  scanf("%d %s", &m, s);
  n = strlen(s);
  for (int i = 0; i < n; i++) v[s[i] - 'a'].push_back(i);
  memset(idx, 0, sizeof(idx));
  int p = 0;
  while (n - p >= m) {
    for (int i = 0; i < 26; i++) {
      while (idx[i] < v[i].size() && v[i][idx[i]] < p) idx[i]++;
      if (idx[i] == v[i].size()) continue;
      if (v[i][idx[i]] >= p + m) continue;
      while (idx[i] + 1 < v[i].size() && v[i][idx[i] + 1] < p + m) idx[i]++;
      p = v[i][idx[i]] + 1;
      w.push_back(i);
      break;
    }
  }
  int c = -1, cnt = 0;
  for (int x : w) {
    if (x > c) {
      c = x;
      cnt = 0;
    }
    if (x == c) cnt++;
  }
  vector<char> ans(cnt, 'a' + c);
  for (int i = 0; i < n; i++)
    if (s[i] - 'a' < c) ans.push_back(s[i]);
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++) s[i] = ans[i];
  s[ans.size()] = 0;
  puts(s);
}
