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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cout << *it << " = " << a << "\n";
  err(++it, args...);
}
const long long N = 2e5 + 5, M = 26, inf = 1e18;
bool ok(vector<int> &a, int m) {
  int cnt = 0;
  for (int i = 0; i < ((int)(a).size()); i++) {
    cnt += a[i] == 0;
    if (a[i] or i + 1 == ((int)(a).size())) {
      if (cnt >= m) return 0;
      cnt = 0;
    }
  }
  return 1;
}
void put(string &s, vector<int> &a, int m, int j, string &ans) {
  vector<int> st;
  int last = -1;
  for (int i = 0; i < ((int)(s).size()); i++) {
    if (s[i] - 'a' == j) {
      st.push_back(i);
    }
    if (a[i]) {
      last = i;
      st.clear();
    }
    if (((int)(st).size()) and i >= last + m) {
      int pos = st.back();
      a[pos] = 1;
      ans += s[pos];
      last = pos;
      st.clear();
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int m;
  string s, ans;
  cin >> m >> s;
  vector<int> a(((int)(s).size()));
  for (int i = 0; i < M; i++) {
    put(s, a, m, i, ans);
    if (ok(a, m)) {
      return cout << ans, 0;
    }
    char c = 'a' + i;
    ans += string(
        count(s.begin(), s.end(), c) - count(ans.begin(), ans.end(), c), c);
  }
}
