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<cstdio>
#include<queue>
#define maxn 500005
using namespace std;
int max(int a,int b){
	return a>b?a:b;
}
int n,k,a[maxn],t[maxn<<2],head[maxn],cnt=0,d[maxn],ans[maxn];
struct edge{
	int nxt,to;
}e[maxn<<1];
priority_queue<int> q;
void change(int p,int l,int r,int L,int R){
	if(l==r){
		t[p]=max(t[p],R);
		return;
	}
	int mid=(l+r)>>1;
	if(L<=mid)
		change(p<<1,l,mid,L,R);
	else
		change(p<<1|1,mid+1,r,L,R);
	t[p]=max(t[p<<1],t[p<<1|1]);
}
int Find(int p,int l,int r,int L,int R){
	if(L<=l&&r<=R){
		return t[p];
	}
	int mid=(l+r)>>1;
	if(R<=mid)
		return Find(p<<1,l,mid,L,R);
	else if(L>mid)
		return Find(p<<1|1,mid+1,r,L,R);
	else
		return max(Find(p<<1,l,mid,L,mid),Find(p<<1|1,mid+1,r,mid+1,R));
}
void add(int u,int v){
	e[++cnt].nxt=head[u];
	e[cnt].to=v;
	head[u]=cnt;
	d[v]++;
}
int main(){
	scanf("%d%d",&n,&k);
	k--;
	for(int i=1;i<=n;i++){
		int x;
		scanf("%d",&x);
		a[x]=i;
	}
	for(int i=1;i<=n;i++){
		int x=Find(1,1,n,a[i],a[i]+k);
		if(x>=1&&x<=n)
			add(a[x],a[i]);
		x=Find(1,1,n,a[i]-k,a[i]);
		if(x>=1&&x<=n)
			add(a[x],a[i]);
		change(1,1,n,a[i],i);
	}
	for(int i=1;i<=n;i++){
		if(!d[i])
			q.push(-i);
	}
	cnt=0;
	while(q.size()){
		int u=-q.top();
		q.pop();
		for(int i=head[u];i;i=e[i].nxt){
			int v=e[i].to;
			if(!(--d[v])){
				q.push(-v);
			}
		}
		ans[u]=++cnt;
	}
	for(int i=1;i<=n;i++)
		printf("%d\n",ans[i]);
	return 0;
}