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>
const int maxn = 1e5 + 10;
using namespace std;
vector<int> a[30];
string s;
bool vis[maxn];
int m;
int main() {
  cin >> m;
  cin >> s;
  int len = s.size();
  for (int i = 0; i <= len - 1; i++) a[s[i] - 'a'].push_back(i);
  int i, mx = 0;
  for (i = 0; i <= len - m; i++) {
    for (int j = 0; j <= 25; j++) {
      int k = upper_bound(a[j].begin(), a[j].end(), i + m - 1) - a[j].begin();
      k--;
      int len = a[j].size();
      if (k >= 0 && k <= len - 1 && a[j][k] >= i && a[j][k] <= i + m - 1) {
        mx = max(mx, j);
        i = a[j][k];
        vis[a[j][k]] = 1;
        break;
      }
    }
  }
  for (int i = 0; i < mx; i++) {
    len = a[i].size();
    for (int j = 0; j <= len - 1; j++) cout << char(i + 'a');
  }
  len = a[mx].size();
  for (int j = 0; j <= len - 1; j++)
    if (vis[a[mx][j]]) cout << char(mx + 'a');
  cout << endl;
  return 0;
}
