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;
#define mid ((l+r)>>1)
const int N=1<<20;
struct node{
	int to,ne;
}e[N];
int n,k,i,mn[N],x,a[N],d[N],tot,h[N],u,v,id[N],cnt;
priority_queue<int,vector<int>,greater<int> >q;
void add(int x,int y){e[++tot]={y,h[x]},h[x]=tot,d[y]++;}
void up(int t){mn[t]=min(mn[t<<1],mn[t<<1|1]);}
int query(int t,int l,int r,int x,int y){
	if (x>y) return 1e9;
	if (x<=l && r<=y) return mn[t];
	int ans=1e9;
	if (x<=mid) ans=min(ans,query(t<<1,l,mid,x,y));
	if (mid<y) ans=min(ans,query(t<<1|1,mid+1,r,x,y));
	return ans;
}
void update(int t,int l,int r,int x,int y){
	if (l==r){
		mn[t]=y;
		return;
	}
	if (x<=mid) update(t<<1,l,mid,x,y);
	else update(t<<1|1,mid+1,r,x,y);
	up(t);
}
int main(){
	scanf("%d%d",&n,&k);
	for (i=1;i<=n;i++) scanf("%d",&x),a[x]=i;
	memset(mn,63,sizeof(mn));
	for (i=n;i;i--){
		x=query(1,1,n,max(a[i]-k+1,1),a[i]-1);
		if (1<=x && x<=n) add(a[i],a[x]);
		x=query(1,1,n,a[i]+1,min(a[i]+k-1,n));
		if (1<=x && x<=n) add(a[i],a[x]);
		update(1,1,n,a[i],i);
	}
	for (i=1;i<=n;i++)
		if (!d[i]) q.push(i);
	while (!q.empty()){
		u=q.top(),q.pop(),id[u]=++cnt;
		for (i=h[u];i;i=e[i].ne)
			if (!(--d[v=e[i].to])) q.push(v);
	}
	for (i=1;i<=n;i++) printf("%d\n",id[i]);
}