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;
int n,tot,k,a[500005],ans[500005],p[500005];
int t[2000005],d[500005],point[1000005],head[500005],nxt[1000005];
priority_queue<int> pq;
void addedge(int x,int y)
{
	d[y]++;
	point[++tot]=y,nxt[tot]=head[x],head[x]=tot;
}
int query(int x,int l,int r,int L,int R)
{
	if(l>R||r<L||L>R||!t[x]) return 0;
	if(l>=L&&r<=R) return t[x];
	int mid=(l+r)>>1;
	return max(query(x<<1,l,mid,L,R),query(x<<1|1,mid+1,r,L,R));
}
void edit(int x,int l,int r,int pos,int val)
{
	t[x]=val;
	if(l==r) return;
	int mid=(l+r)>>1;
	if(pos<=mid) edit(x<<1,l,mid,pos,val);
	else edit(x<<1|1,mid+1,r,pos,val);
}
int main()
{
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++)
	{
		int x;
		scanf("%d",&x);
		p[x]=i;
	}
	for(int i=1;i<=n;i++)
	{
		int pos=query(1,1,n,max(0,p[i]-k+1),p[i]-1);
		if(pos) addedge(p[i],p[pos]);
		pos=query(1,1,n,p[i]+1,min(p[i]+k-1,n));
		if(pos) addedge(p[i],p[pos]);
		edit(1,1,n,p[i],i);
	}
	for(int i=1;i<=n;i++) if(!d[i]) pq.push(i);
	tot=n;
	while(!pq.empty())
	{
		int x=pq.top();
		pq.pop();
		a[tot--]=x;
		for(int i=head[x];i;i=nxt[i])
		{
			int y=point[i];
			d[y]--;
			if(!d[y]) pq.push(y);
		}
	}
	for(int i=1;i<=n;i++) ans[a[i]]=i;
	for(int i=1;i<=n;i++) printf("%d\n",ans[i]);
	return 0;
}