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 n, m;
int vis[100005];
pair<int, int> a[100005];
char s[100005];
bool check() {
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (vis[i])
      cnt = 0;
    else
      cnt++;
    if (cnt >= m) return 0;
  }
  return 1;
}
int calc(int c) {
  for (int i = 1; i <= n; i++)
    if (s[i] - 'a' == c) vis[i] = 0;
  int ret = 0, cnt = 0, cur = 0;
  for (int i = 1; i <= n; i++) {
    if (s[i] - 'a' == c) cur = i;
    if (vis[i])
      cnt = 0, cur = i;
    else
      cnt++;
    if (cnt >= m) ret++, cnt = i - cur, cur = 0;
  }
  return ret;
}
int main() {
  scanf("%d%s", &m, s + 1);
  n = strlen(s + 1);
  for (int i = 1; i <= n; i++) a[i].first = s[i] - 'a', a[i].second = i;
  sort(a + 1, a + 1 + n);
  for (int i = 1; i <= n;) {
    int j = i;
    while (a[i].first == a[j].first && j <= n) j++;
    for (int k = i; k < j; k++) vis[a[k].second] = 1;
    if (check()) {
      int p = calc(a[i].first);
      for (int k = 1; k < i; k++) printf("%c", (char)(a[k].first + 'a'));
      for (int k = 1; k <= p; k++) printf("%c", (char)(a[i].first + 'a'));
      return 0;
    }
    i = j;
  }
  return 0;
}
