You are given a permutation P_1 ... P_N of the set {1, 2, ..., N}.

You can apply the following operation to this permutation, any number of times (possibly zero):

* Choose two indices i,j (1 ≦ i < j ≦ N), such that j - i ≧ K and |P_i - P_j| = 1. Then, swap the values of P_i and P_j.



Among all permutations that can be obtained by applying this operation to the given permutation, find the lexicographically smallest one.

Constraints

* 2≦N≦500,000
* 1≦K≦N-1
* P is a permutation of the set {1, 2, ..., N}.

Input

The input is given from Standard Input in the following format:


N K
P_1 P_2 ... P_N


Output

Print the lexicographically smallest permutation that can be obtained.

Examples

Input

4 2
4 2 3 1


Output

2
1
4
3


Input

5 1
5 4 3 2 1


Output

1
2
3
4
5


Input

8 3
4 5 7 8 3 1 2 6


Output

1
2
6
7
5
3
4
8
#include <bits/stdc++.h>
int main() {
  int N, K;
  scanf("%d %d", &N, &K);
  int P[N];
  for (int i = 0; i < N; i++) {
    scanf("%d", &P[i]);
  }
  int nowmax = N;
  int minsoat[N];
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      if (i + 1 == P[j]) {
        minsoat[i] = j;
        j = N;
      }
    }
  }
  for (int k = 0; k < N; k++) {
    while (nowmax > (1 + k)) {
      if (minsoat[nowmax - 2] - minsoat[nowmax - 1] >= K) {
        do {
          int a = P[minsoat[nowmax - 2]];
          P[minsoat[nowmax - 2]] = P[minsoat[nowmax - 1]];
          P[minsoat[nowmax - 1]] = a;
        } while (0);
        do {
          int a = minsoat[nowmax - 2];
          minsoat[nowmax - 2] = minsoat[nowmax - 1];
          minsoat[nowmax - 1] = a;
        } while (0);
      }
      nowmax = nowmax - 1;
    }
    nowmax = N;
  }
  for (int i = 0; i < N; i++) {
    printf("%d\n", P[i]);
  }
  return 0;
}
