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 inf = 1e9 + 10;
const int maxn = 100005;
int a[maxn];
int b[maxn], dp[maxn];
int n, m;
bool check() {
  for (int i = 1; i <= n; i++) {
    b[i] = b[i - 1] + a[i];
    if (i >= m && b[i] == b[i - m]) {
      return 0;
    }
  }
  return 1;
}
int main() {
  cin >> m;
  string s;
  cin >> s;
  n = (int)s.size();
  vector<int> vec(1, 0);
  for (int i = 'a'; i <= 'z'; i++) {
    for (int j = 1; j <= n; j++) {
      if (s[j - 1] == i) {
        a[j] = 1;
        vec.push_back(j);
      }
    }
    if (check()) {
      sort(vec.begin(), vec.end());
      int st = 0;
      int t = 0;
      for (int j = 1; j < (int)vec.size(); j++) {
        int x = upper_bound(vec.begin() + st, vec.end(), vec[j] - m - 1) -
                vec.begin();
        if (s[vec[j] - 1] != i) {
          t += dp[x];
          st = j;
        } else {
          dp[j] = dp[x] + 1;
        }
      }
      int x = upper_bound(vec.begin() + st, vec.end(), n - m) - vec.begin();
      t += dp[x];
      for (int j = 'a'; j < i; j++) {
        for (int k = 1; k <= n; k++) {
          if (s[k - 1] == j) {
            putchar(j);
          }
        }
      }
      for (int j = 1; j <= t; j++) {
        putchar(i);
      }
      return 0;
    }
  }
}
