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;
void smain();
int main() {
  ios_base::sync_with_stdio(0);
  cout.precision(7);
  cout << fixed;
  smain();
  return 0;
}
int m, n;
char s[200002];
vector<char> res;
bool calc(char x, int l, int r) {
  if (r - l < m) return true;
  vector<int> pos;
  for (int i = l; i < r; ++i)
    if (s[i] == x) pos.push_back(i);
  pos.push_back(r);
  int pr = l - 1, k = (int)pos.size();
  for (int i = 0; i < k; i++) {
    if (pos[i] - pr > m) return false;
    if (i + 1 < k && pos[i + 1] - pr > m) res.push_back(s[pos[i]]), pr = pos[i];
  }
  return true;
}
bool solve(char x) {
  res.clear();
  int pr = -1;
  for (int i = 0; i < n; i++)
    if (s[i] < x) {
      if (!calc(x, pr + 1, i)) return false;
      res.push_back(s[i]);
      pr = i;
    }
  return calc(x, pr + 1, n);
}
void smain() {
  for (; cin >> m >> s;) {
    n = strlen(s);
    for (int i = 0; i < 26; i++)
      if (solve('a' + i)) break;
    sort(res.begin(), res.end());
    for (auto c : res) cout << c;
    cout << endl;
  }
}
