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<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
typedef long long ll;
#define N 1000002
const int p=1e9+7;
struct quji{int l,r,mid,mn;}c[N<<2];
struct node{int x,va;friend bool operator <(node n1,node n2){return n1.va<n2.va;}};
struct edge{int to,nxxt;}e[N<<1];
int n,a[N],b[N],head[N],cnt=1,rd[N],s[N],tot,k;
priority_queue<node>q;
inline void ins(int u,int v){e[cnt].to=v;e[cnt].nxxt=head[u];head[u]=cnt++;}
void bt(int ind,int l,int r)
{
	c[ind].l=l;c[ind].r=r;c[ind].mid=(l+r)>>1;c[ind].mn=1e9;
	if(l==r)return;bt(ind<<1,l,c[ind].mid);bt(ind<<1|1,c[ind].mid+1,r);
}
void chng(int ind,int x,int y)
{
	if(c[ind].l==c[ind].r){c[ind].mn=min(c[ind].mn,y);return;}
	if(x<=c[ind].mid)chng(ind<<1,x,y);else chng(ind<<1|1,x,y);
	c[ind].mn=min(c[ind<<1].mn,c[ind<<1|1].mn);
}
int ask(int ind,int l,int r)
{
	if(l<=c[ind].l&&c[ind].r<=r)return c[ind].mn;
	int ret=1e9;
	if(l<=c[ind].mid)ret=min(ret,ask(ind<<1,l,r));
	if(r>c[ind].mid)ret=min(ret,ask(ind<<1|1,l,r));
	return ret;
}
int main()
{
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++)scanf("%d",&a[i]),b[a[i]]=i;
	for(int i=1,j=n;i<j;i++,j--)swap(b[i],b[j]);
	bt(1,1,n);
	for(int i=n;i;i--)
	{
		int x=ask(1,b[i],min(n,b[i]+k-1));
		if(x<=n)ins(i,x),rd[x]++;
		x=ask(1,max(1,b[i]-k+1),b[i]);
		if(x<=n)ins(i,x),rd[x]++;
		chng(1,b[i],i);
	}
	for(int i=1;i<=n;i++)if(!rd[i])q.push((node){i,b[i]});
	while(!q.empty())
	{
		node a1=q.top();q.pop();s[++tot]=a1.va;;
		for(int i=head[a1.x];i;i=e[i].nxxt)
		{
			int t1=e[i].to;rd[t1]--;if(!rd[t1])q.push((node){t1,b[t1]});
		}
	}
	for(int i=1,j=n;i<j;i++,j--)swap(s[i],s[j]);
	for(int i=1;i<=n;i++)a[s[i]]=i;
	for(int i=1;i<=n;i++)printf("%d\n",a[i]);
}
