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;
vector<long long> Vec[100000];
int pos;
string s;
int M;
bool f(int lim) {
  bool ans = 1, f;
  for (long long i = (0); i < (pos); i++) {
    f = 0;
    for (long long j = (0); j < (lim + 1); j++)
      if (Vec[i][j] > 0) f = 1;
    ans = ans && f;
  }
  return ans;
}
int gg(int v) {
  vector<long long> curr;
  for (long long i = (0); i < (pos); i++) {
    bool f = 0;
    for (long long j = (0); j < (v); j++) {
      if (Vec[i][j] > 0) f = 1;
    }
    if (f == 0) curr.push_back(i);
  }
  int pp = 0, lst = -1, ans = 0;
  while (pp < curr.size()) {
    ans++;
    for (long long i = (curr[pp]); i < (curr[pp] + M); i++)
      if (s[i] - 'a' == v) lst = i;
    while (pp < curr.size() and curr[pp] <= lst) pp++;
  }
  return ans;
}
int main() {
  cin >> M >> s;
  pos = 1;
  int l = 0, r = M;
  vector<long long> v(26, 0);
  for (long long i = (0); i < (M); i++) v[s[i] - 'a']++;
  Vec[0] = v;
  while (pos + M <= s.size()) {
    v[s[l] - 'a']--;
    v[s[r] - 'a']++;
    Vec[pos] = v;
    l++;
    r++;
    pos++;
  }
  int ans;
  string ss = "";
  for (long long i = (0); i < (26); i++) {
    if (f(i)) {
      ans = i;
      break;
    }
  }
  for (long long i = (0); i < (s.size()); i++)
    if (s[i] - 'a' < ans) ss += s[i];
  int cnt = gg(ans);
  for (long long i = (0); i < (cnt); i++) ss += ('a' + ans);
  sort((ss).begin(), (ss).end());
  cout << ss << endl;
}
