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;
set<int> s[26];
int m;
string str;
int ans[26];
set<int> vis;
int main() {
  scanf("%d", &m);
  cin >> str;
  cerr << "siz " << str.size() << endl;
  for (int i = 0; i < (int)str.size(); ++i) s[str[i] - 'a'].insert(i);
  for (int i = m - 1; i < (int)str.size(); ++i) vis.emplace(i);
  for (int i = 0; i < 26; ++i)
    if (!s[i].empty()) {
      for (auto it = vis.begin(); it != vis.end();) {
        auto itt = s[i].upper_bound(*it);
        if (itt == s[i].begin()) {
          ++it;
          continue;
        }
        --itt;
        if (*it - *itt >= m) {
          ++it;
          continue;
        }
        ++ans[i];
        int k = *itt;
        for (; k < min((int)str.size(), *itt + m); ++k) vis.erase(k);
        s[i].erase(itt);
        it = vis.lower_bound(k);
      }
      if (vis.empty())
        break;
      else {
        vector<int> ers;
        ans[i] += s[i].size();
        for (auto j : vis) {
          auto it1 = s[i].upper_bound(j);
          if (it1 != s[i].begin()) {
            --it1;
            if ((j - *it1) < m) ers.push_back(j);
          }
        }
        for (int j : ers) vis.erase(j);
      }
    }
  for (int i = 0; i < 26; ++i) cout << string(ans[i], char('a' + i));
  cout << endl;
}
