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>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int N = 100005;
const int M = 26;
const int inf = 1000000007;
const int mod = 20160519;
char s[N];
int m, v[M];
bool check(int k) {
  int i, p;
  for (i = 0; s[i] != '\0' && s[i] - 'a' > k; i++)
    ;
  if (s[i] == '\0' || i >= m) return false;
  for (p = i + m; s[i] != '\0'; i++) {
    if (s[i] - 'a' <= k) p = i + m;
    if (i >= p) break;
  }
  if (s[i] == '\0') return true;
  return false;
}
int main() {
  int i, k, p, c = 0, pos;
  scanf("%d", &m);
  scanf("%s", s);
  for (i = 0; s[i] != '\0'; i++) v[s[i] - 'a']++;
  for (i = 0; i < 26; i++) {
    if (!v[i]) continue;
    if (check(i)) break;
    while (v[i]--) printf("%c", i + 'a');
  }
  k = i;
  for (i = 0; s[i] != '\0' && s[i] - 'a' >= k; i++)
    if (s[i] - 'a' == k && i < m) pos = i;
  if (s[i] == '\0' || i >= m)
    i = pos, c++, p = pos + m;
  else
    p = i + m;
  for (; s[i] != '\0'; i++) {
    if (s[i] - 'a' < k) p = i + m;
    if (s[i] - 'a' == k) pos = i;
    if (i >= p) i = pos, c++, p = pos + m;
  }
  while (c--) printf("%c", k + 'a');
  puts("");
  return 0;
}
