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 MAX = 1e5 + 10;
vector<int> v[26];
int a[MAX];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int m;
  cin >> m;
  string s;
  cin >> s;
  int n = s.size();
  for (int i = 0; i < n; i++) {
    v[s[i] - 'a'].push_back(i);
    a[i] = 1;
  }
  for (int i = 25; i >= 0; i--) {
    for (int x : v[i]) a[x] = 0;
    int sum = 0;
    bool flag = true;
    for (int j = 0; j < n; j++) {
      sum += a[j];
      if (j >= m) sum -= a[j - m];
      if (j >= m - 1 && sum == 0) {
        flag = false;
        break;
      }
    }
    if (flag) continue;
    sum = 0;
    int idx = 0;
    for (int j = 0; j < n; j++) {
      if (idx < (int)v[i].size() && v[i][idx] == j) idx++;
      sum += a[j];
      if (j >= m) sum -= a[j - m];
      if (j >= m - 1 && sum == 0) {
        sum++;
        a[v[i][idx - 1]] = 1;
      }
    }
    break;
  }
  string ans;
  for (int i = 0; i < n; i++)
    if (a[i]) ans += s[i];
  sort(ans.begin(), ans.end());
  cout << ans;
  return 0;
}
