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;
int k;
string s;
int l[111111], r[111111];
vector<int> pos[26];
void process() {
  for (int i = 0; i < s.size(); ++i) {
    pos[s[i] - 'a'].push_back(i);
    l[i] = r[i] = 0;
  }
}
bool del(int pos) {
  if (l[pos] + r[pos] + 1 >= k) return false;
  int pl = pos - l[pos] - 1;
  int pr = pos + r[pos] + 1;
  if (pr < s.size()) l[pr] += 1 + l[pos];
  if (pl >= 0) r[pl] += 1 + r[pos];
  return true;
}
bool del(char c) {
  bool f = 1;
  while (f && pos[c - 'a'].size() > 0) {
    int pp = pos[c - 'a'].back();
    pos[c - 'a'].pop_back();
    if (del(pp)) f = 0;
  }
  return !f;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> k >> s;
  process();
  string ss = s;
  sort(ss.begin(), ss.end());
  int i;
  for (i = (int)ss.size() - 1; i >= 0; --i) {
    if (!del(ss[i])) break;
  }
  cout << ss.substr(0, i + 1) << "\n";
  return 0;
}
