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 N=5e5+5;
int n,k,pos[N],a[N];
const int INF=0x3f3f3f3f;
struct SegT{
	int t[N<<2];
	void build(int x,int l,int r){
		int mid=(l+r)>>1;
		t[x]=INF;if(l==r)return ;
		build(x<<1,l,mid);build(x<<1|1,mid+1,r);
	}
	void chg(int x,int l,int r,int pos,int sum){
		int mid=(l+r)>>1;
		t[x]=min(t[x],sum);
		if(l==r)return;
		if(pos>mid)chg(x<<1|1,mid+1,r,pos,sum);
		else chg(x<<1,l,mid,pos,sum);
	}
	int qry(int x,int l,int r,int ql,int qr){
		if(ql>qr)return INF;
		int mid=(l+r)>>1,ans=INF;
		if(ql<=l&&qr>=r)return t[x];
		if(ql<=mid)ans=min(ans,qry(x<<1,l,mid,ql,qr));
		if(qr>mid)ans=min(ans,qry(x<<1|1,mid+1,r,ql,qr));
		return ans;
	}
}T;
int hed[N],to[N*3],nxt[N*3],cnt;
int d[N];
inline void adde(int u,int v){
	//cout<<u<<' '<<v<<endl;
	++cnt;to[cnt]=v,nxt[cnt]=hed[u];hed[u]=cnt;
	d[v]++;
}
priority_queue<int,vector<int>,greater<int> >Q;
int ans[N],top;
int main()
{
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++){
		scanf("%d",&a[i]);
		pos[a[i]]=i;
	}
	T.build(1,1,n);
	//for(int i=1;i<=n;i++)cout<<pos[i]<<" ";puts("");
	for(int i=n;i;i--){
		//cout<<i<<" :: ";
		int p=pos[i];
		//cout<<p<<" ? "<<p+1<<" "<<p+k-1<<endl;
		int R=T.qry(1,1,n,p+1,min(n,p+k-1));
		int L=T.qry(1,1,n,max(1,p-k+1),p-1);
		//cout<<L<<" "<<R<<endl;
		if(R<=n)adde(pos[i],pos[R]);
		if(L<=n)adde(pos[i],pos[L]);
		T.chg(1,1,n,pos[i],i);
	}
	for(int i=1;i<=n;i++)if(!d[i]){
		Q.push(i);
		//cout<<"!"<<i<<endl;
	}int top=0;
	while(!Q.empty()){
		int u=Q.top();Q.pop();
		//cout<<u<<endl;
		ans[u]=++top;
		for(int i=hed[u];i;i=nxt[i]){
			int v=to[i];d[v]--;
			if(!d[v])Q.push(v);
		}
	}
	for(int i=1;i<=n;i++)printf("%d\n",ans[i]);
}