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 II() {
  int q;
  scanf("%d", &q);
  return q;
}
long long LII() {
  long long q;
  scanf("%lld", &q);
  return q;
}
void solve() {
  int M;
  cin >> M;
  string s;
  cin >> s;
  int N = (int)s.size();
  deque<int> de[30];
  for (int i = 0; i < N; i++) {
    de[s[i] - 'a'].push_back(i);
  }
  map<int, int> mp;
  for (int i = 0; i < M; i++) mp[s[i] - 'a']++;
  string ss = "";
  int prev_mi = 100, prev_in;
  for (int i = 0; i < 26; i++) {
    if (mp[i] > 0) {
      prev_mi = i;
      ss += (char)(i + 'a');
      break;
    }
  }
  for (int i = M - 1; i >= 0; i--) {
    if ((s[i] - 'a') == prev_mi) {
      prev_in = i;
      break;
    }
  }
  de[s[0] - 'a'].pop_front();
  for (int i = 1; i < N; i++) {
    int q = i + M - 1;
    if (q >= N) break;
    mp[s[i - 1] - 'a']--;
    mp[s[i + M - 1] - 'a']++;
    int mi = 100;
    for (int j = 0; j < 26; j++) {
      if (mp[j] > 0) {
        mi = j;
        break;
      }
    }
    if (mi == prev_mi) {
      if (prev_in >= i) continue;
      int temp = 0;
      while (de[mi].size() > 0 && de[mi].front() <= q) {
        temp = de[mi].front();
        de[mi].pop_front();
      }
      prev_in = temp;
      ss += (char)(mi + 'a');
    } else {
      int temp = 0;
      while (de[mi].size() > 0 && de[mi].front() <= q) {
        temp = de[mi].front();
        de[mi].pop_front();
      }
      prev_in = temp;
      prev_mi = mi;
      ss += (mi + 'a');
    }
  }
  sort(ss.begin(), ss.end());
  char ch = ss.back();
  map<char, int> mp1;
  string s1 = "";
  for (auto it : s) mp1[it]++;
  for (char cc = 'a'; cc < ch; cc++) {
    int q = mp1[cc];
    while (q--) s1 += cc;
  }
  for (auto it : ss)
    if (it == ss.back()) s1 += it;
  cout << s1 << endl;
}
int32_t main() {
  int tt;
  tt = 1;
  int cas = 1;
  while (tt--) {
    solve();
  }
}
