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 = (int)1010;
const int INF = (int)2e9 + 10;
const long double EPS = (long double)1e-8;
const long long MOD = (long long)1e9 + 7;
const int ALPH = 26;
bool normal(set<int> const& s, int m) {
  auto cur = s.begin();
  ++cur;
  while (cur != s.end()) {
    auto prev = cur;
    --prev;
    if (*cur - *prev > m) {
      return false;
    }
    ++cur;
  }
  return true;
}
void solve() {
  vector<int> let[26];
  int m;
  string s;
  cin >> m >> s;
  int n = s.size();
  for (int i = 0; i < n; ++i) {
    let[s[i] - 'a'].push_back(i);
  }
  set<int> pos({-1, n});
  string ret = "";
  for (int c = 0; c < 26; ++c) {
    for (int i = 0; i < (int)let[c].size(); ++i) {
      int ind = let[c][i];
      int next = n;
      if (i < (int)let[c].size() - 1) {
        next = let[c][i + 1];
      }
      auto iter = pos.upper_bound(ind);
      next = min(next, *iter);
      --iter;
      int prev = *iter;
      if (next - prev > m) {
        ret += ('a' + c);
        pos.insert(ind);
      }
    }
    if (normal(pos, m)) {
      break;
    } else {
      for (auto& x : let[c]) {
        if (pos.find(x) == pos.end()) {
          pos.insert(x);
          ret += ('a' + c);
        }
      }
    }
  }
  cout << ret << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  solve();
  return 0;
}
