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 long long N = (long long)(5e5) + 322;
const long long INF = (long long)(1e9);
const long long mod = (long long)(1e9) + 7;
const double eps = 1e-9;
string s, ans, id;
int m, used[N], col[N], pre[N], n;
char ch;
pair<char, int> t[N];
inline int solve() {
  int res = 0, last = -1, lastch = -1;
  for (long long i = 0; i <= (int)(s.size()) - 1; ++i) {
    if (s[i] < ch) last = i;
    if (s[i] == ch) lastch = i;
    if (i - m >= last) {
      if (i - m >= lastch) return 0;
      res++;
      last = lastch;
    }
  }
  return res;
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  cin >> m >> s;
  for (ch = 'a'; ch <= 'z'; ch++) {
    int cnt = solve();
    if (solve() != 0) {
      for (long long i = 1; i <= cnt; ++i) ans += ch;
      break;
    }
    for (long long i = 0; i <= (int)(s.size()) - 1; ++i)
      if (s[i] == ch) ans += ch;
  }
  cout << ans;
  return 0;
}
