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;
const int N = int(1e5 + 5);
vector<char> used;
int n, m, ans[26], all[26];
char s[N];
bool solve(int c) {
  char cc = char('a' + c);
  int l = 0, last = -1;
  for (int i = 0; i < m - 1; ++i) {
    if (s[i] == cc) {
      last = i;
    }
  };
  for (int r = m - 1; r < n;) {
    if (s[r] == cc) {
      last = r;
    }
    if (used[r]) {
      ++r;
      ++l;
      continue;
    }
    if (last >= l) {
      int q = last;
      ans[c]++;
      while (r < n && l <= q) {
        used[r] = true;
        if (s[r] == cc) {
          last = r;
        }
        ++r;
        ++l;
      }
    } else {
      ans[c] = all[c];
      for (; r < n; ++r, ++l) {
        if (s[r] == cc) {
          last = r;
        }
        if (last >= l) {
          used[r] = true;
        }
      }
      return false;
    }
  }
  return true;
}
int main(int argc, char *argv[]) {
  scanf("%d", &m);
  scanf("%s", &s);
  n = int(strlen(s));
  used.resize(n, false);
  memset(ans, 0, sizeof(ans));
  memset(all, 0, sizeof(all));
  for (int i = 0; s[i]; ++i) {
    all[(s[i] - 'a')]++;
  }
  for (int i = 0; !solve(i); ++i)
    ;
  for (int i = 0; i < 26; ++i) {
    while (ans[i]--) {
      printf("%c", char('a' + i));
    }
  }
  return 0;
}
