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;
struct node {
  int x, v;
  bool operator<(const node &a) const { return v > a.v; }
};
char S[100010];
char ansv[100010];
int n, m, tot, f[100010];
priority_queue<node> q;
int main() {
  scanf("%d%s", &m, S);
  n = strlen(S);
  for (char x = 'a'; x <= 'z'; x++) {
    for (int i = 1; i <= n; i++) f[i] = 0x3f3f3f3f;
    f[0] = 0;
    while (!q.empty()) q.pop();
    q.push((node){0, 0});
    int cnt = 0;
    for (int i = 1; i <= n; i++) {
      if (S[i - 1] > x) continue;
      if (S[i - 1] == x) cnt++;
      while (!q.empty()) {
        node tmp = q.top();
        if (tmp.x < i - m)
          q.pop();
        else {
          if (S[i - 1] == x)
            f[i] = tmp.v + 1;
          else
            f[i] = tmp.v;
          break;
        }
      }
      if (f[i] < 0x3f3f3f3f) q.push((node){i, f[i]});
    }
    int ans = 0x3f3f3f3f;
    for (int i = n - m + 1; i <= n; i++) ans = min(ans, f[i]);
    if (ans < 0x3f3f3f3f) {
      while (ans--) ansv[++tot] = x;
      break;
    } else
      while (cnt--) ansv[++tot] = x;
  }
  for (int i = 1; i <= tot; i++) putchar(ansv[i]);
  putchar('\n');
  return 0;
}
