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 MAXN = 1e5 + 5;
int m, ans[26];
char s[MAXN];
bool vis[MAXN];
pair<int, int> cnt[26];
int main() {
  scanf("%d%s", &m, s);
  for (int i = 0; i < m; i++)
    cnt[s[i] - 'a'].first++, cnt[s[i] - 'a'].second = i;
  for (int l = 0, r = m - 1, len = strlen(s), pos = -1; r < len;) {
    if (l <= pos) {
      if (r == len - 1) break;
      cnt[s[l++] - 'a'].first--;
      cnt[s[++r] - 'a'].first++, cnt[s[r] - 'a'].second = r;
      continue;
    }
    for (int i = 0; i < 26; i++)
      if (cnt[i].first) {
        ans[i]++, pos = cnt[i].second, vis[pos] = 1;
        break;
      }
    if (r == len - 1) break;
    cnt[s[l++] - 'a'].first--;
    cnt[s[++r] - 'a'].first++, cnt[s[r] - 'a'].second = r;
  }
  int flag;
  for (int i = 25; i >= 0; i--)
    if (ans[i]) {
      flag = i;
      break;
    }
  for (int i = 0, len = strlen(s); i < len; i++)
    if (!vis[i] && (s[i] - 'a' < flag)) ans[s[i] - 'a']++;
  for (int i = 0; i < 26; i++)
    while (ans[i]) putchar('a' + i), ans[i]--;
  return 0;
}
