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>
struct read_item {
  read_item(){};
  template <class T>
  operator T() {
    T t;
    std::cin >> t;
    return t;
  }
};
char splf(int i, int n) { return (i + 1 < n) ? ' ' : '\n'; };
struct RI {
  int i;
  RI(int a) : i(a) {}
  int operator*() { return i; }
  void operator++() { i += 1; }
  bool operator!=(const RI& r) { return i != r.i; }
};
struct range {
  RI x, y;
  RI begin() { return x; }
  RI end() { return y; }
  range(int a, int b) : x(a), y(b) {}
  range(int a) : x(0), y(a) {}
};
using namespace std;
int main() {
  int m = read_item();
  string s = read_item();
  int len = s.size();
  vector<bool> used(len, false);
  for (auto lim : range(26)) {
    fill(used.begin(), used.end(), false);
    bool ok = true;
    int prev = -1;
    int candidate = -1;
    for (auto i : range(len)) {
      int c = s[i] - 'a';
      if (c <= lim - 1) {
        prev = i;
        candidate = -1;
        used[i] = true;
        continue;
      }
      if (c == lim) {
        candidate = i;
      }
      if (prev + m == i) {
        if (candidate >= 0) {
          used[candidate] = true;
          prev = candidate;
          candidate = -1;
          continue;
        }
        ok = false;
        break;
      }
    }
    if (ok) {
      string ans;
      for (auto i : range(len)) {
        if (used[i]) {
          ans.push_back(s[i]);
        }
      }
      sort(ans.begin(), ans.end());
      printf("%s\n", ans.c_str());
      return 0;
    }
  }
  return 0;
}
