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>
char str[100010], out[100010], visited[100010];
int n, m, len, ar[100010], counter[256], tree[100010];
void update(int p, int v) {
  p++;
  while (p <= n) {
    tree[p] += v;
    p += (p & (-p));
  }
}
int query(int p) {
  int res = 0;
  while (p) {
    res += tree[p];
    p ^= (p & (-p));
  }
  return res;
}
int range_query(int l, int r) {
  l++, r++;
  if (l > r) return 0;
  return (query(r) - query(--l));
}
int main() {
  int i, j, k, l, c, res, flag;
  while (scanf("%d", &m) != EOF) {
    scanf("%s", str);
    n = strlen(str);
    memset(counter, 0, sizeof(counter));
    memset(tree, 0, sizeof(tree));
    for (i = 0; i < n; i++) counter[str[i]]++;
    for (c = 'a'; c <= 'z'; c++) {
      for (i = 0; i < n; i++) {
        if (str[i] == c) update(i, 1);
      }
      flag = 1;
      for (i = 0; (i + m) <= n; i++) {
        if (range_query(i, i + m - 1) == 0) {
          flag = 0;
          break;
        }
      }
      if (flag) break;
    }
    for (i = 0, len = 0; i < n; i++) {
      if (str[i] == c) {
        update(i, -1);
        ar[len++] = i;
      }
    }
    res = 0;
    memset(visited, 0, sizeof(visited));
    for (i = 0, j = 0; (i + m) <= n; i++) {
      if (range_query(i, i + m - 1) == 0) {
        k = -1;
        while (j < len) {
          if ((ar[j] + m - 1) < i)
            j++;
          else {
            if (ar[j] > (i + m - 1)) break;
            k = j++;
          }
        }
        assert(k != -1);
        visited[k] = 1;
        update(ar[k], 1);
      }
    }
    for (i = 0; i < 100010; i++) res += visited[i];
    len = 0;
    for (i = 'a'; i < c; i++) {
      for (j = 0; j < counter[i]; j++) {
        out[len++] = i;
      }
    }
    for (i = 0; i < res; i++) out[len++] = c;
    out[len] = 0;
    puts(out);
  }
  return 0;
}
