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 MAXN = 1e5 + 5;
char s[MAXN];
int main() {
  int m;
  scanf("%d", &m);
  scanf("%s", s);
  int n = strlen(s);
  priority_queue<int, vector<int>, greater<int> > Q;
  for (int k = 0; k < 26; k++) {
    while (!Q.empty()) Q.pop();
    bool flag;
    for (int i = 0; i <= n - 1; i++) {
      flag = 0;
      int tmp = 0;
      for (int j = i; j <= i + m - 1 && j < n; j++) {
        if (s[j] - 'a' < k) Q.push(s[j]), flag = 1, tmp = j;
      }
      if (i > n - m) {
        flag = 1;
        break;
      }
      if (!flag) {
        for (int j = i + m - 1; j >= i; j--) {
          if (s[j] - 'a' == k) {
            Q.push(s[j]), flag = 1;
            tmp = j;
            break;
          }
        }
        if (!flag) break;
      }
      i = tmp;
    }
    if (flag) {
      while (!Q.empty()) {
        printf("%c", Q.top());
        Q.pop();
      }
      break;
    }
  }
  return 0;
}
