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;
int len;
char str[100005];
vector<int> pos[26];
bool vis[100005];
int N = 100000;
int a[100005];
int lowbit(int x) { return x & (-x); }
void add(int x, int info) {
  for (int i = x; i <= N; i += lowbit(i)) a[i] += info;
}
int sum(int x) {
  int ans = 0;
  for (int i = x; i > 0; i -= lowbit(i)) ans += a[i];
  return ans;
}
bool check(int ps) {
  for (int i = 0; i < pos[ps].size(); i++) add(pos[ps][i], 1);
  bool flag = true;
  for (int i = m; i <= len; i++) {
    if (sum(i) - sum(i - m) == 0) {
      flag = false;
      break;
    }
  }
  if (flag)
    for (int i = 0; i < pos[ps].size(); i++) add(pos[ps][i], -1);
  return flag;
}
void solve(int ps) {
  int siz = pos[ps].size();
  for (int i = m; i <= len; i++) {
    if (sum(i) - sum(i - m) == 0) {
      printf("%c", 'a' + ps);
      int pps =
          upper_bound(pos[ps].begin(), pos[ps].end(), i) - pos[ps].begin() - 1;
      add(pos[ps][pps], 1);
    }
  }
}
int main() {
  scanf("%d", &m);
  scanf("%s", str + 1);
  len = strlen(str + 1);
  for (int i = 1; i <= len; i++) {
    pos[str[i] - 'a'].push_back(i);
  }
  for (int i = 0; i < 26; i++) {
    if (check(i)) {
      solve(i);
      return 0;
    }
    for (int j = 0; j < pos[i].size(); j++) printf("%c", 'a' + i);
  }
  printf("\n");
  return 0;
}
