Problem:
Score : 2000 points

Problem Statement

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.

Sample Input 1

4 2
4 2 3 1

Sample Output 1

2
1
4
3

One possible way to obtain the lexicographically smallest permutation is shown below:

4 2 3 1

4 1 3 2

3 1 4 2

2 1 4 3

Sample Input 2

5 1
5 4 3 2 1

Sample Output 2

1
2
3
4
5

Sample Input 3

8 3
4 5 7 8 3 1 2 6

Sample Output 3

1
2
6
7
5
3
4
8

Solution in C++ (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define oo 500005
using namespace std;
int n, a[oo], p[oo], 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;
}