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 N = 1e5 + 5;
const int INF = 1e9 + 5;
int k;
set<int> pos[300];
string s, ans;
bool used[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> k;
  cin >> s;
  s = '#' + s;
  int n = (int)s.size() - 1;
  for (int i = 1; i <= n; ++i) {
    pos[(s[i] - 0)].insert(i);
  }
  int i = 1;
  while (i + k - 1 <= n) {
    int l = i;
    int r = i + k - 1;
    for (int ch = 'a'; ch <= 'z'; ++ch) {
      auto it = pos[ch].lower_bound(l);
      if (it != pos[ch].end() && *it <= r) {
        auto it2 = pos[ch].upper_bound(r);
        --it2;
        ans.push_back(ch);
        i = *it2 + 1;
        break;
      }
    }
  }
  sort(ans.begin(), ans.end());
  for (int i = 'a'; i < ans.back(); ++i) {
    for (int cnt = 1; cnt <= (int)pos[i].size(); ++cnt) {
      cout << char(i);
    }
  }
  sort(ans.begin(), ans.end());
  for (char ch : ans)
    if (ans.back() == ch) cout << ch;
  return 0;
}
