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;
bool used[100010];
int tr[400010];
string str, ans;
int m;
void build(int node, int l, int r) {
  if (l == r)
    tr[node] = l;
  else {
    build(node * 2, l, (l + r) / 2);
    build(node * 2 + 1, (l + r) / 2 + 1, r);
    int p1 = tr[node * 2], p2 = tr[node * 2 + 1];
    if (str[p1] == str[p2])
      tr[node] = max(p1, p2);
    else
      tr[node] = (str[p1] > str[p2] ? p2 : p1);
  }
}
int query(int node, int l, int r, int i, int j) {
  if (i > r || j < l || l > r) return 0x3f3f3f3f;
  if (l >= i && r <= j) return tr[node];
  int p1 = query(node * 2, l, (l + r) / 2, i, j);
  int p2 = query(node * 2 + 1, (l + r) / 2 + 1, r, i, j);
  if (p1 == 0x3f3f3f3f) return p2;
  if (p2 == 0x3f3f3f3f) return p1;
  if (str[p1] == str[p2]) return max(p1, p2);
  if (str[p1] > str[p2]) return p2;
  return p1;
}
int main() {
  cin >> m;
  cin >> str;
  build(1, 0, str.size());
  char maxchar = 0;
  for (int i = 0; i < str.size();) {
    int q = query(1, 0, str.size(), i, min((int)str.size() - 1, i + m - 1));
    ans.push_back(str[q]);
    used[q] = true;
    maxchar = max(maxchar, str[q]);
    if (q + m >= str.size()) break;
    i = q + 1;
  }
  for (int i = 0; i < str.size(); i++) {
    if (str[i] < maxchar && !used[i]) {
      ans.push_back(str[i]);
    }
  }
  sort(ans.begin(), ans.end());
  cout << ans << endl;
  return 0;
}
