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;
const int INF=1e7+5;
priority_queue<int,vector<int>,greater<int> > q;
int n,k;
int A[maxn],ans[maxn],ind[maxn];
int tot;
int h[maxn],last[maxn<<1],to[maxn<<1];
int st[maxn<<2];
void build(int x,int l,int r)
{
	st[x]=INF;
	if(l==r) return;
	int mid=(l+r)>>1;
	build(x<<1,l,mid),build(x<<1|1,mid+1,r);
} 
void pushup(int x)
{
	st[x]=min(st[x<<1],st[x<<1|1]);
}
void update(int x,int l,int r,int id,int k)
{
	if(l==r){
		st[x]=k;
		return;
	}
	int mid=(l+r)>>1;
	if(id<=mid) update(x<<1,l,mid,id,k);
	else update(x<<1|1,mid+1,r,id,k);
	pushup(x);
}
int query(int x,int l,int r,int ql,int qr)
{
	if(r<ql||l>qr) return INF;
	if(l>=ql&&r<=qr) return st[x];
	int mid=(l+r)>>1;
	return min(query(x<<1,l,mid,ql,qr),query(x<<1|1,mid+1,r,ql,qr));
}
void add(int x,int y)
{
	to[++tot]=y;last[tot]=h[x];h[x]=tot;++ind[y];
}
int main()
{
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;++i){
		int a=0;
		scanf("%d",&a);
		A[a]=i;
	}
	build(1,1,n);
	for(int i=n;i>=1;--i){
		int l=query(1,1,n,A[i],A[i]+k-1),r=query(1,1,n,A[i]-k+1,A[i]);
		if(l!=INF) add(A[i],A[l]);
		if(r!=INF) add(A[i],A[r]);
		update(1,1,n,A[i],i);
	}
	for(int i=1;i<=n;++i) if(!ind[i]) q.push(i);
	int cnt=0;
	while(q.size()){
		int u=q.top();
		q.pop();
		ans[u]=++cnt; 
		for(int i=h[u];i;i=last[i]){
			int v=to[i];
			if(!--ind[v]) q.push(v);
		}
	}
	for(int i=1;i<=n;++i) printf("%d\n",ans[i]);
	return 0;
}