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 = 801000;
int a[M], n, m, lenb, f[M];
char s[M];
struct node {
  int val, pl;
} b[M];
bool cmp(const node &x, const node &y) {
  if (x.val == y.val) return x.pl < y.pl;
  return x.val < y.val;
}
bool deal(char x) {
  for (int i = 1; i <= n; i++)
    if (b[i].val == x) a[b[i].pl] = 1;
  int ans = 0, k = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] == 0)
      f[i] = f[i - 1] + 1;
    else
      f[i] = 0;
    ans = max(ans, f[i]);
  }
  return ans < m;
}
void solve(char x) {
  int st = 0, hh;
  for (int i = 1; i <= n; i++) {
    if (b[i].val > x) break;
    if (b[i].val == x) {
      hh = i;
      break;
    }
    printf("%c", (char)b[i].val);
  }
  int num = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] == 1)
      st = i;
    else if (i - st >= m) {
      while (b[hh].pl < i && b[hh + 1].val == x) hh++;
      if (b[hh].pl > i) hh--;
      num++;
      st = b[hh].pl;
    }
  }
  for (int i = 1; i <= num; i++) printf("%c", x);
  return;
}
int main() {
  scanf("%d%s", &m, s + 1);
  n = strlen(s + 1);
  for (int i = 1; i <= n; i++) {
    b[i].val = s[i];
    b[i].pl = i;
  }
  sort(b + 1, b + 1 + n, cmp);
  int k = 0;
  for (int i = 'a'; i <= 'z'; i++)
    if (deal(i)) {
      k = i - 1;
      break;
    }
  for (int i = 1; i <= n; i++) a[i] = 0;
  for (int i = 'a'; i <= k; i++) deal(i);
  solve(k + 1);
  return 0;
}
