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 inf = (int)1e9;
const int mod = (int)1e9 + 7;
int m, n;
string second;
vector<int> g[26];
bool taken[100100];
void take_all(int c) {
  for (int i = 0; i < g[c].size(); i++) {
    taken[g[c][i]] = true;
  }
}
bool is_enough() {
  int last_pos = -1;
  for (int i = 0; i < n; i++) {
    if (taken[i]) {
      if (last_pos == -1) {
        if (i >= m) return false;
      } else if (last_pos + m < i) {
        return false;
      }
      last_pos = i;
    }
  }
  return last_pos != -1 && last_pos + m >= n;
}
void remove_extra(int c) {
  vector<int> v;
  for (int i = 0; i < n; i++) {
    if (taken[i]) v.push_back(i);
  }
  vector<int> prefix;
  for (int i = 0; i < (int)v.size(); i++) {
    int prev_pos = ((prefix.empty()) ? -1 : prefix.back());
    int next_pos = ((i == (int)v.size() - 1) ? n : v[i + 1]);
    if (second[v[i]] - 'a' == c && prev_pos + m >= next_pos) {
      taken[v[i]] = false;
    } else {
      prefix.push_back(v[i]);
    }
  }
}
string collect_taken() {
  string ans = "";
  for (int i = 0; i < n; i++) {
    if (taken[i]) ans += second[i];
  }
  return ans;
}
int main() {
  cin >> m;
  cin >> second;
  n = (int)second.size();
  for (int i = 0; i < second.length(); i++) {
    g[second[i] - 'a'].push_back(i);
  }
  for (int c = 0; c < 26; c++) {
    take_all(c);
    if (is_enough()) {
      remove_extra(c);
      break;
    }
  }
  assert(is_enough());
  string ans = collect_taken();
  sort(ans.begin(), ans.end());
  cout << ans << endl;
  return 0;
}
