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;
char s[1 << 20] = {0};
int main() {
  int m, cnt[26] = {0};
  scanf("%d%s", &m, s);
  int n = strlen(s);
  vector<vector<int>> p(26);
  for (int i = n - 1; i >= 0; i--) p[s[i] - 'a'].push_back(i);
  set<int> S;
  S.insert(-1);
  S.insert(n);
  for (int k = 0; k < 26; k++) {
    auto it = S.begin(), jt = ++S.begin();
    for (int i = p[k].size() - 1; i >= 0 && *it < n; it = jt++) {
      if (*jt - *it > m) {
        for (; i >= 0 && p[k][i] <= *it; i--)
          ;
        for (; i > 0 && p[k][i - 1] <= *it + m; i--)
          ;
        if (i >= 0 && *it < p[k][i] && p[k][i] < *jt) {
          cnt[k]++;
          jt = S.insert(p[k][i]).first;
        }
      }
    }
  }
  for (int i = 25; i >= 0; i--)
    if (cnt[i] > 0) {
      for (int j = i - 1; j >= 0; j--) cnt[j] = p[j].size();
      break;
    }
  for (int i = 0; i < 26; i++)
    for (; cnt[i] > 0; cnt[i]--) putchar('a' + i);
  putchar('\n');
}
