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>
using namespace std;
int n, k, F = false;
int num[500010], ans[500010];
void work1() {
  F = true;
  memset(ans, 0, sizeof(ans));
  for (int i = 1; i <= k; i++) ans[i] = num[i];
  for (int i = k + 1; i <= n; i++) {
    ans[i] = num[i];
    for (int j = 1; j <= i - k; j++) {
      if (ans[j] - 1 == ans[i]) {
        F = false;
        int t = ans[i];
        ans[i] = ans[j];
        ans[j] = t;
        j = 0;
      }
    }
  }
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) scanf("%d", &num[i]);
  if (k == 1) {
    for (int i = 1; i <= n; i++) cout << i << endl;
    return 0;
  }
  if (k == n - 1) {
    if (num[1] > num[n]) {
      int t = num[1];
      num[1] = num[n];
      num[n] = t;
    }
    for (int i = 1; i <= n; i++) cout << num[i] << endl;
    return 0;
  }
  while (!F) {
    work1();
    for (int i = 1; i <= n; i++) num[i] = ans[i];
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << endl;
  return 0;
}
