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
/*
  mail: mleautomaton@foxmail.com
  author: MLEAutoMaton
  This Code is made by MLEAutoMaton
*/
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<iostream>
using namespace std;
#define ll long long
#define REP(a,b,c) for(int a=b;a<=c;a++)
#define re register
#define file(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout)
inline int gi(){
	int f=1,sum=0;char ch=getchar();
	while(ch>'9' || ch<'0'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0' && ch<='9'){sum=(sum<<3)+(sum<<1)+ch-'0';ch=getchar();}
	return f*sum;
}
const int N=1000010;
int p[N],a[N],n,k,d[N],tot,ans[N];
vector<int>G[N];
priority_queue<int,vector<int>,greater<int> >Q;
void Add(int u,int v){G[u].push_back(v);d[v]++;}
int t[N<<1];
void modify(int o,int l,int r,int pos,int v){
	if(l==r){t[o]=v;return;}
	int mid=(l+r)>>1;
	if(pos<=mid)modify(o<<1,l,mid,pos,v);
	else modify(o<<1|1,mid+1,r,pos,v);
	t[o]=min(t[o<<1],t[o<<1|1]);
}
int query(int o,int l,int r,int posl,int posr){
	if(posl>posr)return t[0];
	if(posl<=l && r<=posr)return t[o];
	int mid=(l+r)>>1,ret=t[0];
	if(posl<=mid)ret=min(ret,query(o<<1,l,mid,posl,posr));
	if(mid<posr)ret=min(ret,query(o<<1|1,mid+1,r,posl,posr));
	return ret;
}
int main(){
	n=gi();k=gi();
	for(int i=1;i<=n;i++)p[i]=gi(),a[p[i]]=i;
	memset(t,127/3,sizeof(t));
	for(int i=n;i;i--){
		int x=query(1,1,n,a[i]+1,min(n,a[i]+k-1));
		if(x<t[0])Add(a[i],a[x]);
		x=query(1,1,n,max(1,a[i]-k+1),a[i]-1);
		if(x<t[0])Add(a[i],a[x]);
		modify(1,1,n,a[i],i);
	}
	for(int i=1;i<=n;i++)if(!d[i])Q.push(i);
	while(!Q.empty()){
		int u=Q.top();Q.pop();ans[u]=++tot;
		for(int v:G[u]){d[v]--;if(!d[v])Q.push(v);}
	}
	for(int i=1;i<=n;i++)printf("%d\n",ans[i]);
	return 0;
}
