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;
int a[233333], b[233333];
struct node {
  int val;
  node *next;
} * head[233333], *tail[233333], pool[233333];
set<int> bl;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    b[a[i]] = i;
  }
  head[0] = tail[0] = &pool[0];
  for (int i = 1; i <= n; i++) {
    head[i] = tail[i] = &pool[i];
    head[i]->val = b[i];
  }
  bl.insert(0);
  for (int i = 1; i <= k; i++) bl.insert(a[i]);
  for (int i = 1; i <= n; i++) {
    bl.erase(a[i]);
    auto pos = *--bl.lower_bound(a[i]);
    tail[pos]->next = head[a[i]];
    tail[pos] = tail[a[i]];
    if (i + k <= n) bl.insert(a[i + k]);
  }
  int top = 0;
  for (node *u = head[0]; u; u = u->next) {
    b[top++] = u->val;
  }
  for (int i = 1; i <= n; i++) a[b[i]] = i;
  for (int i = 1; i <= n; i++) cout << a[i] << endl;
  return 0;
}
