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_MAX = 1e5;
int N, M;
char a[N_MAX + 5];
int cnt[26];
int nxt[N_MAX + 5][26];
int ansnxt[N_MAX + 5];
string ans;
int Next(int i, int j) { return min(ansnxt[j + 1], nxt[j + 1][i]); }
int main() {
  ios::sync_with_stdio(false);
  cin >> M >> (a + 1);
  N = strlen(a + 1);
  for (int i = 1; i <= N; ++i) {
    ++cnt[a[i] - 'a'];
    ansnxt[i] = N + 1;
  }
  ansnxt[0] = ansnxt[N + 1] = N + 1;
  for (int i = 0; i < 26; ++i) nxt[N + 1][i] = N + 1;
  for (int i = N; i > 0; --i) {
    memcpy(nxt[i], nxt[i + 1], sizeof nxt[i]);
    nxt[i][a[i] - 'a'] = i;
  }
  memcpy(nxt[0], nxt[1], sizeof nxt[0]);
  for (int i = 0; i < 26; ++i) {
    bool ok = true;
    int j = 0, nr = 0, last = 0;
    while (j <= N) {
      int nx = Next(i, j);
      if (a[nx] != i + 'a') last = nx;
      if (nx - j > M)
        ok = false;
      else if (a[nx] == i + 'a' && Next(i, nx) - last > M) {
        ++nr;
        last = nx;
      }
      while (j < nx) ansnxt[++j] = nx;
    }
    ansnxt[0] = ansnxt[1];
    if (ok) {
      for (int j = 0; j < nr; ++j) ans += 'a' + i;
      break;
    } else {
      for (int j = 0; j < cnt[i]; ++j) ans += 'a' + i;
    }
  }
  cout << ans << "\n";
  return 0;
}
