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;
const int inf = 1e9 + 7;
int a[500010], b[500010];
int sm[500010];
int k;
void sort(int s, int e) {
  if (e - s < 2) return;
  int m = s + e >> 1;
  sort(s, m);
  sort(m, e);
  int p = s, q = m;
  sm[m] = inf;
  int i;
  for (i = m; i-- > s;) sm[i] = min(a[i], sm[i + 1]);
  for (i = s; i < e; i++) {
    if (p >= m)
      b[i] = a[q++];
    else if (q >= e)
      b[i] = a[p++];
    else if (sm[p] >= a[m] + k)
      b[i] = a[q++];
    else
      b[i] = a[p++];
  }
  for (i = s; i < e; i++) a[i] = b[i];
  return;
}
int n, i;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> k;
  for (i = 0; i < n; i++) {
    int s;
    cin >> s;
    a[s - 1] = i;
  }
  sort(0, n);
  for (i = 0; i < n; i++) b[a[i]] = i + 1;
  for (i = 0; i < n; i++) cout << b[i] << '\n';
  cout << endl;
  return 0;
}
