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 + 100;
const int MOD = 1e9 + 7;
const int seed = 1331;
char A[maxn];
char C[maxn];
int B[maxn];
int main() {
  int n, m;
  m = 0;
  int i, j, k;
  scanf("%d", &n);
  scanf("%s", A + 1);
  int l = strlen(A + 1);
  int a = 'z' + 1, b;
  for (i = 'a'; i <= 'z'; i++) {
    m = 0;
    for (j = 1; j <= l; j++)
      if (A[j] <= i) B[m++] = j;
    bool flag = true;
    if (m == 0 || B[0] > n || l - n >= B[m - 1]) flag = false;
    for (j = 0; j < m - 1 && flag; j++)
      if (B[j + 1] - B[j] > n) {
        flag = false;
        break;
      }
    if (flag) break;
  }
  m = 0;
  for (j = 1; j <= l; j++)
    if (A[j] <= i) C[m] = A[j], B[++m] = j;
  k = 0;
  for (j = 1; j <= m; j++) {
    if (A[B[j]] == i) {
      int c = l + 1;
      if (j + 1 <= m) c = B[j + 1];
      int cc = 0;
      if (k != 0) cc = B[k];
      if (c - cc <= n) continue;
    }
    B[++k] = B[j];
  }
  m = k;
  for (i = 1; i <= m; i++) C[i - 1] = A[B[i]];
  C[m] = '\0';
  sort(C, C + m);
  puts(C);
  return 0;
}
