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, ans[30];
char s[100010];
int tree[100010 * 4];
vector<int> adj[30];
void Build(int i, int l, int r) {
  if (l == r) {
    tree[i] = r - l + 1;
    return;
  }
  int mid = (l + r) >> 1;
  Build(i * 2, l, mid);
  Build(i * 2 + 1, mid + 1, r);
  tree[i] = tree[i * 2] + tree[i * 2 + 1];
}
void Modify(int i, int l, int r, int s, int t) {
  if (!tree[i]) return;
  if (s <= l && r <= t) {
    tree[i] = 0;
    return;
  }
  int mid = (l + r) >> 1;
  if (l <= t && mid >= s) Modify(i * 2, l, mid, s, t);
  if (mid < t && r >= s) Modify(i * 2 + 1, mid + 1, r, s, t);
  tree[i] = tree[i * 2] + tree[i * 2 + 1];
}
int Query(int i, int l, int r, int s, int t) {
  if (!tree[i]) return 0;
  if (s <= l && r <= t) return tree[i];
  int mid = (l + r) >> 1, res1 = 0, res2 = 0;
  if (l <= t && mid >= s) res1 = Query(i * 2, l, mid, s, t);
  if (mid < t && r >= s) res2 = Query(i * 2 + 1, mid + 1, r, s, t);
  return res1 + res2;
}
int main() {
  int i, j, k, cnt;
  scanf("%d%s", &m, s + 1);
  n = strlen(s + 1);
  for (i = 1; i <= n; ++i) adj[s[i] - 'a'].push_back(i);
  n = n - m + 1;
  Build(1, 1, n);
  for (i = 0; i < 26; ++i) {
    cnt = 0;
    for (k = 0; k < adj[i].size(); ++k) {
      if (!Query(1, 1, n, max(1, adj[i][k] - m + 1), min(adj[i][k], n)))
        continue;
      if (k + 1 < adj[i].size() &&
          (max(1, adj[i][k] - m + 1) >= max(1, adj[i][k + 1] - m + 1) ||
           !Query(1, 1, n, max(1, adj[i][k] - m + 1),
                  max(1, adj[i][k + 1] - m + 1) - 1)))
        continue;
      Modify(1, 1, n, max(1, adj[i][k] - m + 1), min(adj[i][k], n));
      ++cnt;
    }
    if (cnt)
      for (j = 0; j < i; ++j) ans[j] = adj[j].size();
    ans[i] = cnt;
  }
  for (i = 0; i < 26; ++i)
    for (j = 1; j <= ans[i]; ++j) printf("%c", 'a' + i);
  return 0;
}
