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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<set>
#include<vector>
#include<queue>
#define SF scanf
#define PF printf
#define MAXN 500010
using namespace std;
int p[MAXN],q[MAXN];
set<int> pre,bac;
vector<int> a[MAXN];
priority_queue<int> que;
int n,k,pri[MAXN];
int vis[MAXN];
int main(){
	SF("%d%d",&n,&k);
	for(int i=1;i<=n;i++){
		SF("%d",&p[i]);
		q[p[i]]=i;
	}
	for(int i=1;i<=k;i++)
		bac.insert(p[i]);
	set<int>::iterator b,g;
	for(int i=1;i<=n;i++){
		b=bac.upper_bound(p[i]);
		g=pre.upper_bound(p[i]);
		if(b!=bac.end()){
			//PF("{%d %d}\n",q[*b],i);
			a[q[*b]].push_back(i);
			vis[i]++;
		}
		if(g!=pre.end()){
			//PF("[%d %d]\n",q[*g],i);
			a[q[*g]].push_back(i);
			vis[i]++;
		}
		pre.insert(p[i]);
		if(i-k+1>0)
			pre.erase(p[i-k+1]);
		if(i+k<=n)
			bac.insert(p[i+k]);
		bac.erase(p[i]);
	}
	for(int i=1;i<=n;i++)
		if(vis[i]==0)
			que.push(i);
	int tot=0;
	while(!que.empty()){
		int x=que.top();
		que.pop();
		pri[++tot]=x;
		//PF("[%d]",x);
		for(int i=0;i<a[x].size();i++){
			vis[a[x][i]]--;
			if(vis[a[x][i]]==0)
				que.push(a[x][i]);
		}
	}
	for(int i=1;i<=n;i++)
		q[pri[i]]=n-i+1;
	for(int i=1;i<=n;i++)
		PF("%d\n",q[i]);
}