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
/*
一个长度为n的排列，每次可交换 |i−j|≥K 并且 |a_i−a_j|=1 的数对, 求经过若干次变换后最小字典序的排列
*/
#include<set>
#include<iostream>
#include<cstdio>
using namespace std;

const int N=500010;

struct node{
	int s,t,nxt;
}e[2*N];

int n,k,tot,p[N],q[N],h[N],d[N];

set<int>S;
set<int> :: iterator it;

int main(){
//	freopen("permutation.in","r",stdin);
//	freopen("permutation.out","w",stdout);
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++)
		scanf("%d",&p[i]),q[p[i]]=i;
	for(int i=1;i<=n;i++){
		if(i>k) S.erase(p[i-k]);
		it=S.insert(p[i]).first;
		if ((++it)!=S.end()){
			e[++tot]=(node){i,q[*it],h[i]};
			h[i]=tot;d[q[*it]]++;
		}
	}
	S.clear();
	for(int i=n;i;i--){
		if(i+k<=n) S.erase(p[i+k]);
		it=S.insert(p[i]).first;
		if((++it)!=S.end()){
			e[++tot]=(node){i,q[*it],h[i]};
			h[i]=tot;d[q[*it]]++;
		}
	}
	S.clear();
	for(int i=1;i<=n;i++)
		if(!d[i]) S.insert(i);
	for(int i=1;i<=n;i++){
		it=S.begin(),S.erase(it),p[*it]=i;
		for(int j=h[*it];j;j=e[j].nxt)
			if (!(--d[e[j].t]))
				S.insert(e[j].t);
	}
	for(int i=1;i<=n;i++)
		printf("%d\n",p[i]);
	return 0;
}