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[500010];
int p[500010];
void file() {
  freopen("permutation.in", "r", stdin);
  freopen("permutation.out", "w", stdout);
}
void solve(int now, int i) {
  int hh = p[a[now] - 1];
  if (hh > i) return;
  while ((hh - now) >= k && hh && hh < i) {
    p[a[now] - 1] = now;
    p[a[now]] = hh;
    swap(a[hh], a[now]);
    hh = p[a[now] - 1];
  }
}
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 = k + 1; i <= n; i++) {
    int no = p[a[i] + 1];
    while ((i - no) >= k && no) {
      p[a[i] + 1] = i;
      p[a[i]] = no;
      swap(a[no], a[i]);
      solve(no, i);
      no = p[a[i] + 1];
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%d\n", a[i]);
  }
  return 0;
}
