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 = 100010;
char str[N];
pair<int, int> range[N];
int m;
set<int> s;
bool apaga(int k) {
  s.erase(k);
  if (s.empty()) return false;
  set<int>::iterator prev = s.lower_bound(k);
  if (prev == s.begin()) return range[*prev].first == range[k].first;
  prev--;
  set<int>::iterator prox = s.lower_bound(k);
  if (prox == s.end()) return range[*prev].second == range[k].second;
  return range[*prev].second + 1 >= range[*prox].first;
}
int main(void) {
  scanf("%d %s", &m, str);
  int sz = strlen(str);
  vector<pair<int, int> > v;
  for (int i = 0; i < sz; i++) {
    v.emplace_back(str[i], i);
    s.insert(i);
  }
  int x = 0, y = m - 1;
  for (int i = 0; i <= sz - m; i++) {
    range[x].second = i;
    x++;
    y++;
    range[y].first = i + 1;
  }
  while (x < sz) range[x++].second = sz - m;
  sort(v.rbegin(), v.rend());
  for (int i = 0; i < v.size(); i++) {
    if (apaga(v[i].second)) continue;
    int at = v[i].first;
    while (i < v.size() and v[i].first == at) {
      if (!apaga(v[i].second)) s.insert(v[i].second);
      i++;
    }
    break;
  }
  string ans;
  for (auto c : s) {
    ans.push_back(str[c]);
  }
  sort(ans.begin(), ans.end());
  cout << ans << endl;
  return 0;
}
