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>
static const int MAXN = 100005;
int n = 0, m, ans_len = 0;
char s[MAXN], ans[MAXN] = {'\0'};
std::pair<char, int> p[MAXN];
std::set<int> marked;
std::multiset<int> distance;
int main() {
  scanf("%d", &m);
  getchar();
  while ((s[n] = getchar()) >= 'a' && s[n] <= 'z') ++n;
  for (int i = 0; i < n; ++i) p[i] = std::make_pair(s[i], i);
  std::sort(p, p + n);
  marked.insert(-1);
  marked.insert(n);
  distance.insert(n + 1);
  for (char c = 'a'; c <= 'z'; ++c) {
    for (int i = 0; i < n; ++i)
      if (p[i].first == c) {
        std::set<int>::iterator prev = marked.lower_bound(p[i].second),
                                succ = prev--;
        int j = i + 1;
        while (j < n && p[j].first == c && p[j].second < *succ) ++j;
        int last_marked = *prev;
        if (*succ <= last_marked + m) continue;
        for (int k = i; k < j; ++k) {
          while (k < j && p[k].second <= last_marked + m) ++k;
          --k;
          if (k < i || p[k].second == last_marked) break;
          prev = marked.lower_bound(p[k].second), succ = prev--;
          ans[ans_len++] = p[k].first;
          marked.insert(p[k].second);
          distance.erase(distance.find(*succ - *prev));
          distance.insert(*succ - p[k].second);
          distance.insert(p[k].second - *prev);
          last_marked = p[k].second;
          if (*succ <= last_marked + m) break;
        }
        i = j - 1;
      }
    if (*(distance.rbegin()) <= m) break;
    for (int i = 0; i < n; ++i)
      if (p[i].first == c) {
        if (marked.find(p[i].second) != marked.end()) continue;
        std::set<int>::iterator prev = marked.lower_bound(p[i].second),
                                succ = prev--;
        ans[ans_len++] = p[i].first;
        marked.insert(p[i].second);
        distance.erase(distance.find(*succ - *prev));
        distance.insert(*succ - p[i].second);
        distance.insert(p[i].second - *prev);
      }
  }
  puts(ans);
  return 0;
}
