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 NMAX = 100000;
int cnt[27];
int kekokek;
set<int> used[27];
int main() {
  srand((unsigned int)time(NULL));
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int m;
  cin >> m;
  string s;
  cin >> s;
  int n = (int)s.length();
  int L = 0, R = 0;
  vector<pair<int, int> > a;
  while (R < m) {
    int c = s[R] - 'a';
    used[c].insert(R);
    cnt[c]++;
    R++;
  }
  R--;
  for (int i = 0; i < 26; i++) {
    if (cnt[i]) {
      a.push_back({i, *used[i].rbegin()});
      break;
    }
  }
  while (R < n - 1) {
    int c = s[L] - 'a';
    cnt[c]--;
    used[c].erase(L);
    L++;
    R++;
    c = s[R] - 'a';
    cnt[c]++;
    used[c].insert(R);
    for (int i = 0; i < 26; i++) {
      if (cnt[i]) {
        int last = *used[i].rbegin();
        pair<int, int> x = {i, *used[i].begin()};
        if (x.first < a.back().first) {
          a.push_back({x.first, last});
        } else if (x.first == a.back().first) {
          if (a.back().second < x.second) {
            a.push_back({x.first, last});
          }
        } else if (x.first > a.back().first) {
          a.push_back({x.first, last});
        }
        break;
      }
    }
  }
  sort(a.begin(), a.end());
  int val = 0;
  for (int i = 1; i < (int)a.size(); i++) {
    if (a[i].first != a[i - 1].first) val++;
  }
  sort(s.begin(), s.end());
  int last = a.back().first;
  for (int i = 0; i < n; i++) {
    if (s[i] < last + 'a') {
      cout << s[i];
    }
  }
  for (int i = 0; i < a.size(); i++) {
    if (a[i].first == last) {
      cout << char(last + 'a');
    }
  }
  return 0;
}
