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;
map<char, int> tmp;
string ans;
const int maxn = 1e5 + 100;
bool vis[maxn];
int main() {
  ans.clear();
  long long n;
  string str;
  cin >> n;
  cin >> str;
  int len = str.size() - n;
  int flag = 0;
  memset(vis, 0, sizeof(vis));
  for (int i = 0; i < n; ++i) {
    tmp[str[i]]++;
  }
  int pre = 0;
  map<char, int>::iterator it = tmp.begin();
  char m;
  for (int i = n - 1; i >= 0; --i) {
    if (str[i] == it->first) {
      flag = i;
      m = str[i];
      vis[i] = 1;
      break;
    }
  }
  for (int i = 1; i < len; ++i) {
    tmp[str[i - 1]]--;
    tmp[str[i + n - 1]]++;
    char c;
    for (it = tmp.begin(); it != tmp.end(); ++it) {
      if (tmp[it->first]) {
        c = it->first;
        break;
      }
    }
    if (c == m) {
      if (flag < i) {
        for (int j = i + n - 1; j >= i; --j) {
          if (str[j] == c) {
            vis[j] = 1;
            flag = j;
            break;
          }
        }
      }
    } else if (c > m) {
      m = c;
      for (int j = i + n - 1; j >= i; --j) {
        if (str[j] == c) {
          vis[j] = 1;
          flag = j;
          break;
        }
      }
    }
  }
  for (int i = 0; i < str.size(); ++i) {
    if (str[i] < m)
      ans.push_back(str[i]);
    else if (str[i] == m && vis[i])
      ans.push_back(str[i]);
  }
  sort(ans.begin(), ans.end());
  cout << ans << endl;
  return 0;
}
