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 M = 200001;
char s[M];
char ans[M];
int vis[M];
int main() {
  int m;
  scanf("%d %s", &m, s + 1);
  int n = strlen(s + 1);
  int len = 0;
  for (int i = 0; i < 26; i++) {
    for (int j = 1; j <= n; j++) vis[j] = 0;
    for (int j = 1; j <= n; j++)
      if (s[j] <= 'a' + i) vis[j] = 1;
    int lv = 0;
    bool ok = 1;
    for (int j = 1; j <= n; j++) {
      if (vis[j])
        lv = j;
      else if (j - lv >= m) {
        ok = 0;
        break;
      }
    }
    if (!ok) continue;
    for (int j = 1; j <= n; j++) vis[j] = 0;
    for (int j = 1; j <= n; j++)
      if (s[j] < 'a' + i) {
        ans[len++] = s[j];
        vis[j] = 1;
      }
    lv = 0;
    int last;
    for (int j = 1; j <= n; j++) {
      if (vis[j]) lv = j;
      if (s[j] == 'a' + i) last = j;
      if (j == lv + m) {
        lv = last;
        ans[len++] = 'a' + i;
      }
    }
    sort(ans, ans + len);
    cout << ans << endl;
    return 0;
  }
}
