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;
int m;
char s[101000];
bool is[101000];
bool p[101000];
vector<char> v;
int n;
bool judge(int x) {
  for (int i = 1; i <= n; i++)
    if (s[i] < x + 'a') is[i] = true;
  int yuan;
  int wh = 0;
  int you = 0;
  for (int i = m; i; i--) {
    if (is[i]) {
      wh = i;
      break;
    }
  }
  if (!wh) {
    for (int i = m; i; i--) {
      if (s[i] == x + 'a') {
        wh = i;
        break;
      }
    }
  }
  if (!wh) return false;
  p[wh] = true;
  yuan = wh + m - 1;
  for (int i = wh + 1; i <= n; i++) {
    if (is[i]) {
      yuan = i + m - 1;
    }
    if (s[i] == 'a' + x) you = i;
    if (i > yuan) {
      if (you + m - 1 >= i) {
        p[you] = true;
        yuan = you + m - 1;
      } else {
        return false;
      }
    }
  }
  return true;
}
void print(int x) {
  for (int i = 1; i <= n; i++) {
    if (s[i] < x + 'a' || p[i] == 1) {
      v.push_back(s[i]);
    }
  }
  sort(v.begin(), v.end());
  for (int i = 0; i < (int)v.size(); i++) printf("%c", v[i]);
  puts("");
}
int main() {
  scanf("%d", &m);
  scanf("%s", s + 1);
  n = strlen(s + 1);
  for (int i = 0; i < 26; i++) {
    if (judge(i)) {
      print(i);
      return 0;
    }
  }
  return 0;
}
