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;

const int MaxN=5e5+5;

int N,K,M;
int P[MaxN],Pos[MaxN];
int Deg[MaxN],Q[MaxN];

set<int> S;
set<int>::iterator it;
vector<int> To[MaxN];
priority_queue<int,vector<int>,greater<int> > H;

void Add_Edge(int u,int v){
	To[u].push_back(v);
	Deg[v]++;
}

int main(){
	int i,u,v;
	scanf("%d%d",&N,&K);
	for(i=1;i<=N;i++){
		scanf("%d",&P[i]);
		Pos[P[i]]=i;
	}
	for(i=2;i<=N;i++){
		S.insert(P[i-1]);
		if(i>K)
			S.erase(P[i-K]);
		if((it=S.upper_bound(P[i]))!=S.end())
			Add_Edge(i,Pos[*it]);
	}
	S.clear();
	for(i=N-1;i;i--){
		S.insert(P[i+1]);
		if(i+K<=N)
			S.erase(P[i+K]);
		if((it=S.upper_bound(P[i]))!=S.end())
			Add_Edge(i,Pos[*it]);
	}
	for(i=1;i<=N;i++)
		if(!Deg[i]){
			H.push(i);
			//printf("%d ",i);
		}
	//puts("");
	while(H.size()){
		u=H.top();
		Q[u]=++M;
		H.pop();
		for(i=0;i<To[u].size();i++)
			if(!--Deg[v=To[u][i]])
				H.push(v);
	}
	for(i=1;i<=N;i++)
		printf("%d\n",Q[i]); 
	return 0;
}