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, a[500005], p[500005], k;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]), p[a[i]] = i;
  for (int i = 1; i <= n; ++i) {
    while (a[i] != 1) {
      if (p[a[i] - 1] - p[a[i]] >= k) {
        int p1 = p[a[i] - 1];
        swap(p[a[i] - 1], p[a[i]]);
        swap(a[i], a[p1]);
        continue;
      }
      if (a[i] > 2 && p[a[i] - 1] > p[a[i]] && p[a[i] - 2] - p[a[i] - 1] >= k) {
        printf("%d %d\n", i, a[i]);
        int p1 = p[a[i] - 1], p2 = p[a[i] - 2];
        swap(p[a[i] - 1], p[a[i] - 2]);
        swap(a[p1], a[p2]);
        p1 = p[a[i] - 1];
        swap(p[a[i] - 1], p[a[i]]);
        swap(a[i], a[p1]);
        continue;
      }
      break;
    }
  }
  for (int i = 1; i <= n; ++i) printf("%d\n", a[i]);
  return 0;
}
