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 main() {
  int t = 1;
  while (t--) {
    int n;
    cin >> n;
    string a;
    cin >> a;
    deque<pair<char, int> > d;
    d.push_back({a[0], 0});
    vector<int> pos[26];
    for (int i = 0; a[i]; i++) {
      pos[a[i] - 'a'].push_back(i);
    }
    for (int i = 1; i < n; i++) {
      while (!d.empty() && d.back().first >= a[i]) {
        d.pop_back();
      }
      d.push_back({a[i], i});
    }
    vector<char> mi;
    mi.push_back(d.front().first);
    for (int i = n; a[i]; i++) {
      int start = i - n + 1;
      while (!d.empty() && d.front().second < start) d.pop_front();
      while (!d.empty() && d.back().first > a[i]) {
        d.pop_back();
      }
      d.push_back({a[i], i});
      mi.push_back(d.front().first);
    }
    int l = a.length();
    set<int> po;
    for (int i = 0; i + n - 1 < l; i++) {
      char mm = mi[i];
      int last;
      if (po.size()) last = *(--po.end());
      if (po.size() && a[last] == mm && last >= i) continue;
      int position =
          (upper_bound(pos[mm - 'a'].begin(), pos[mm - 'a'].end(), i + n - 1) -
           pos[mm - 'a'].begin()) -
          1;
      position = pos[mm - 'a'][position];
      po.insert(position);
    }
    char m = 'z';
    for (char i : a) m = min(i, m);
    char ma = 'a';
    for (int i : po) ma = max(a[i], ma);
    set<char> it;
    for (int i : po) it.insert(a[i]);
    if (it.size() != 1) {
      for (int i = 0; a[i]; i++) {
        if ((a[i] < ma || a[i] == m) && po.count(i) == 0) po.insert(i);
      }
    }
    vector<int> s;
    for (int i : po) {
      s.push_back(a[i]);
    }
    sort(s.begin(), s.end());
    for (char i : s) cout << i;
  }
}
