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;
int n, m, sum[1005];
char a[100005];
inline int check(char x) {
  int ans = 0, last = 0, p = -1;
  for (int i = 'a'; i < x; i++) ans += sum[i];
  for (int i = 1; i <= n; i++) {
    if (a[i] < x) last = i;
    if (a[i] == x) p = i;
    if (i - last == m)
      if (p > last)
        last = p, ans++;
      else
        return -1;
  }
  return ans;
}
int main() {
  scanf("%d%s", &m, a + 1);
  n = strlen(a + 1);
  for (int i = 1; i <= n; i++) sum[a[i]]++;
  int ans = -1;
  for (int i = 'a'; i <= 'z'; i++)
    if (ans == -1) ans = check(i);
  for (int i = 'a'; i <= 'z'; i++)
    if (sum[i] <= ans) {
      for (int j = 1; j <= sum[i]; j++) cout << char(i);
      ans -= sum[i];
    } else {
      for (int j = 1; j <= ans; j++) cout << char(i);
      return 0;
    }
  return 0;
}
