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 arr[30];
map<char, int> mp;
set<char> res;
set<char> exist;
int main(int argc, char** argv) {
  int m;
  cin >> m;
  string s;
  cin >> s;
  int size = s.size();
  for (int i = 0; i < size; i++) {
    arr[s[i] - 'a']++;
    exist.insert(s[i]);
  }
  int l = 0;
  int r = m - 1;
  for (int i = 0; i <= r; i++) {
    mp[s[i]]++;
  }
  if (m == size) {
    cout << mp.begin()->first;
    return 0;
  }
  auto it = mp.begin();
  res.insert(it->first);
  exist.erase(it->first);
  while (r < size - 1) {
    if (mp[s[l]] == 1) {
      mp.erase(s[l]);
    } else
      mp[s[l]]--;
    l++;
    r++;
    mp[s[r]]++;
    char toAdd = mp.begin()->first;
    auto first = exist.begin();
    while (*first < toAdd && first != exist.end()) {
      res.insert(*first);
      exist.erase(first);
      first = exist.begin();
    }
    res.insert(mp.begin()->first);
  }
  string ans = "";
  int counter = 0;
  int limit = res.size();
  for (auto it : res) {
    if (counter != limit - 1) {
      for (int i = 0; i < arr[it - 'a']; i++) {
        ans += it;
      }
    } else {
    }
    counter++;
  }
  int index = -1;
  int amount = 0;
  char last = *res.rbegin();
  int lastB = 0;
  res.erase(last);
  for (int i = 0; i < m; i++) {
    if (res.find(s[i]) != res.end()) index = i;
    if (s[i] == last) lastB = i;
  }
  if (index == -1) {
    amount++;
    index = lastB;
  }
  l = 0;
  r = m - 1;
  while (r < size - 1) {
    l++;
    r++;
    if (res.find(s[r]) != res.end()) {
      index = r;
    }
    if (s[r] == last) lastB = r;
    if (index == l - 1) {
      amount++;
      index = lastB;
    }
  }
  for (int i = 0; i < amount; i++) {
    ans += last;
  }
  cout << ans;
  return 0;
}
