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>
const long long mod = 100000007;
const long long M = 100005;
const double pi = acos(-1);
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  int m;
  string second;
  cin >> m;
  cin >> second;
  second = '0' + second + '}';
  int n = ((int)(second).size()), pos[n];
  int cnt[n][26];
  memset(cnt, 0, sizeof(cnt));
  set<pair<char, int> > z;
  pos[0] = -1;
  z.insert(make_pair(second[0], 0));
  for (int i = (1); i < (n); i++) {
    pos[i] = z.begin()->second;
    z.insert(make_pair(second[i], i));
    if (i >= m) z.erase(make_pair(second[i - m], i - m));
  }
  int cur = n - 1;
  string t;
  while (cur != -1) {
    t += second[cur];
    second[cur] = '{';
    cur = pos[cur];
  }
  sort((t).begin(), (t).end());
  t = t.substr(1, ((int)(t).size()) - 2);
  string p = "";
  for (int i = (1); i < (n - 1); i++)
    if (second[i] < t[((int)(t).size()) - 1]) p += second[i];
  t = t + p;
  sort((t).begin(), (t).end());
  cout << t << "\n";
  return 0;
}
