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 double PI = acos(-1.0);
const int INF = 1000 * 1000 * 1000 + 7;
const long long LINF = 1LL * INF * INF;
const int MAX = 20004;
int m;
string s;
multiset<pair<char, int> > q;
string solve(char up) {
  int p = -1;
  string res = "";
  int lastUp = -1;
  for (int i = (0); i < ((int)s.size() + 1); ++i) {
    if (i - p > m) {
      if (lastUp > p && i - lastUp <= m) {
        res += s[lastUp];
        p = lastUp;
      } else {
        return "";
      }
    }
    if (i == (int)s.size()) break;
    if (s[i] < up) {
      res += s[i];
      p = i;
    }
    if (s[i] == up) lastUp = i;
  }
  sort(res.begin(), res.end());
  return res;
}
int main() {
  cin >> m >> s;
  for (int up = ('a'); up < ('z' + 1); ++up) {
    string ans = solve(up);
    if (ans != "") {
      cout << ans;
      return 0;
    }
  }
  return 0;
}
