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, a[505000], bel[505000], L[1010], R[1010], mn[1010], an[1010],
    pos[1010], sz = 500, tot, ans[505000], inf = 1e8;
void RE(int w) {
  int now = inf, res = inf, p;
  for (int i = L[w]; i <= R[w]; ++i) {
    if (now - a[i] >= k) res = min(res, a[i]), p = i;
    now = min(now, a[i]);
  }
  mn[w] = now, an[w] = res, pos[w] = p;
}
int doit() {
  int now = inf, res = inf, p;
  for (int i = 1; i <= tot; ++i) {
    if (now - mn[i] >= k) res = min(res, mn[i]), p = i;
    now = min(now, mn[i]);
  }
  return pos[p];
}
int main() {
  cin >> n >> k;
  for (int i = 1, x; i <= n; ++i)
    scanf("%d", &x), a[x] = i, tot = bel[i] = (i - 1) / sz + 1;
  for (int i = 1; i <= tot; ++i)
    L[i] = i * sz - sz + 1, R[i] = min(n, i * sz), RE(i);
  for (int i = 1, p; i <= n; ++i)
    ans[a[p = doit()]] = i, a[p] = inf, RE(bel[p]);
  for (int i = 1; i <= n; ++i) printf("%d\n", ans[i]);
}
