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;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long m;
  cin >> m;
  string s;
  cin >> s;
  vector<long long> was(26);
  for (auto it : s) {
    was[it - 'a']++;
  }
  pair<long long, long long> lol;
  for (long long i = 0; i < 26; i++) {
    vector<bool> used((long long)(s).size());
    vector<long long> nxt_can((long long)(s).size());
    nxt_can[(long long)(s).size() - 1] = 1e9;
    long long kek = 1e9;
    for (long long j = (long long)(s).size() - 1; j >= 0; j--) {
      if (s[j] - 'a' < i) {
        used[j] = true;
      }
      nxt_can[j] = kek;
      if (s[j] - 'a' <= i) {
        kek = j;
      }
    }
    long long cnt = 0, ptr = 0;
    long long lst = -1;
    bool ok = true;
    for (long long j = 0; j < (long long)(s).size(); j++) {
      if (used[j]) {
        lst = j;
        if ((long long)(s).size() - 1 - lst <= m) {
          break;
        }
      } else if (s[j] - 'a' == i && nxt_can[j] - lst > m) {
        cnt++;
        lst = j;
        ptr++;
        if ((long long)(s).size() - 1 - lst <= m) {
          break;
        }
      } else if (j - lst >= m) {
        ok = false;
      }
    }
    if (ok) {
      lol = {i, cnt};
      break;
    }
  }
  for (long long i = 0; i < lol.first; i++) {
    for (long long j = 0; j < was[i]; j++) {
      cout << char('a' + i);
    }
  }
  for (long long j = 0; j < lol.second; j++) {
    cout << char('a' + lol.first);
  }
  cout << '\n';
}
