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;
using ll = long long;
const int NAX = 2e5 + 5, MOD = 1000000007;
void solveCase() {
  int m;
  cin >> m;
  string str;
  cin >> str;
  int n = str.size();
  vector<vector<int>> cnt(n, vector<int>(26));
  vector<int> poses[26];
  for (size_t i = 0; i < n; i++) {
    poses[str[i] - 'a'].push_back(i);
    if (i > 0) cnt[i] = cnt[i - 1];
    cnt[i][str[i] - 'a']++;
  }
  for (int c = 0; c < 26; c++) {
    if (poses[c].empty()) continue;
    string res;
    int now = 0;
    int prev = -1;
    bool golbOk = true;
    while (now < n) {
      int cc = str[now] - 'a';
      ;
      if (cc < c) {
        ;
        prev = now;
        ++now;
      } else if ((now - prev) >= m) {
        ;
        int j = now;
        int cnt = m;
        bool ok = false;
        while (j >= 0 && cnt > 0) {
          if (str[j] == ('a' + c)) {
            res.push_back(str[j]);
            ok = true;
            prev = j;
            break;
          }
          --cnt;
          --j;
        }
        ++now;
        if (!ok) {
          golbOk = false;
          break;
        }
      } else
        now++;
    }
    if (golbOk) {
      string res2;
      for (int c1 = 0; c1 < c; c1++)
        for (size_t i = 0; i < poses[c1].size(); i++) res2.push_back(c1 + 'a');
      ;
      cout << res2 << res << '\n';
      return;
    }
  }
  assert(false);
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  for (int i = 1; i <= t; ++i) solveCase();
  return 0;
}
