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;
vector<unsigned> W[26];
set<unsigned> S;
int main() {
  ios_base::sync_with_stdio(0);
  int m;
  string s;
  cin >> m >> s;
  for (unsigned i = 0; i < s.length(); ++i) W[s[i] - 'a'].push_back(i);
  int T[26] = {0};
  unsigned cur;
  bool add;
  for (int i = 0; i < 26; ++i) {
    cur = 0;
    auto x = S.lower_bound(m);
    auto j = lower_bound(W[i].begin(), W[i].end(), m);
    if (x != S.begin()) x--;
    if (j != W[i].begin()) j--;
    add = false;
    while (cur <= s.length() - m) {
      if (x != S.end() && *x < cur + m) {
        cur = *x + 1;
      } else if (j != W[i].end() && *j < cur + m) {
        cur = *j + 1;
        S.insert(*j);
      } else {
        add = true;
        break;
      }
      x = S.lower_bound(cur + m);
      j = lower_bound(W[i].begin(), W[i].end(), cur + m);
      if (x != S.begin() && *(prev(x)) >= cur) x--;
      if (j != W[i].begin() && *(j - 1) >= cur) j--;
    }
    if (add) {
      for (auto c : W[i]) S.insert(c);
    } else if (cur > s.length() - m)
      break;
  }
  for (auto x : S) T[s[x] - 'a']++;
  for (int i = 0; i < 26; ++i)
    while (T[i]--) cout << char('a' + i);
  return 0;
}
