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 M 500005
int n,k,cnt;
int a[M],p[M],deg[M],ans[M],minn[M<<2];

vector <int> vec[M];
priority_queue <int> q;

void tpsort(){
	for(int i=1; i<=n; i++)
		if(!deg[i]) q.push(-i);
	while(!q.empty()){
		int x=-q.top(); q.pop();
		a[++cnt]=x;
		for(int i=0; i<vec[x].size(); i++)
			if(--deg[vec[x][i]]==0) q.push(-vec[x][i]);
	}
}

void build(int k,int l,int r){
	minn[k]=0x3f3f3f;
	if(l==r) return ;
	int mid=(l+r)>>1;
	build(k<<1,l,mid); 
	build(k<<1|1,mid+1,r);
} 

void modify(int k,int l,int r,int x,int v){
	if(l==r) { minn[k]=min(minn[k],v);return ;}
	int mid=(l+r)>>1;
	if(x<=mid) modify(k<<1,l,mid,x,v);
	if(x>mid) modify(k<<1|1,mid+1,r,x,v);
	minn[k]=min(minn[k<<1],minn[k<<1|1]);
}

int query(int k,int l,int r,int x,int y){
	if(l>=x&&r<=y) return minn[k];
	int mid=(l+r)>>1;
	int res=0x3f3f3f;
	if(x<=mid) res=min(res,query(k<<1,l,mid,x,y));
	if(y>mid) res=min(res,query(k<<1|1,mid+1,r,x,y));
	return res;
}
void d(){ puts("HH");}
int main()
{
	cin >> n >> k;
	for(int i=1; i<=n; i++) scanf("%d",&a[i]), p[a[i]]=i;
	build(1,1,n);
	for(int i=n;i ;i--){ // 倒序 （并不知道为什么
		int t1=query(1,1,n,max(1,p[i]-k+1),p[i]);
		if(t1<=n) vec[p[i]].push_back(p[t1]), deg[p[t1]]++;
		int t2=query(1,1,n,p[i],min(n,p[i]+k-1));
		if(t2<=n) vec[p[i]].push_back(p[t2]), deg[p[t2]]++;
	//	cout<<i<<" "<<t1<<" "<<t2<<endl;
		modify(1,1,n,p[i],i); // 在p[i]位置加入i 
	}
	//for(int i=1; i<=n; i++) cout<<deg[i]<<" ";
	tpsort();  //拓扑排序 
	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;
}