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;
char s[100010];
int ans[30];
int val[100010];
bool check(int x) {
  int i, now = 0;
  for (i = 1; i <= n; i++)
    if (s[i] - 'a' == x) val[max(1, i - m + 1)]++, val[i + 1]--;
  for (i = 1; i + m - 1 <= n; i++) {
    now += val[i];
    if (!now) goto FAILED;
  }
  for (i = 1; i <= n; i++)
    if (s[i] - 'a' == x) val[max(1, i - m + 1)]--, val[i + 1]++;
  return true;
FAILED:
  return false;
}
int findNext(int goal, int x, int bgn) {
  int i, ret = bgn;
  for (i = bgn + 1; i <= n; i++)
    if (s[i] - 'a' == x) {
      if (i - m + 1 > goal) break;
      ret = i;
    }
  return ret;
}
int work(int x) {
  int i, j, now = 0, ret = 0;
  for (i = j = 1; i + m - 1 <= n; i++) {
    now += val[i];
    if (now) continue;
    j = findNext(i, x, j);
    now++;
    ret++;
    val[max(1, j - m + 1)]++;
    val[j + 1]--;
  }
  return ret;
}
int main() {
  int i, j, x;
  scanf("%d%s", &m, s + 1);
  n = strlen(s + 1);
  for (i = 0; i < 26; i++)
    if (check(i)) break;
  x = i;
  for (i = 1; i <= n; i++)
    if (s[i] - 'a' < x) ans[s[i] - 'a']++;
  ans[x] = work(x);
  for (i = 0; i <= x; i++)
    for (j = 1; j <= ans[i]; j++) printf("%c", i + 'a');
  return 0;
}
