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<cstring>
#include<vector>
#include<queue>
#include<set>
#include<cstdio>
#include<algorithm>
using namespace std;

#define LL long long

inline int read(){
	int x=0,f=1;char c=getchar();
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) x=x*10+c-'0';
	return x*f;
}
const int INF=9999999;
const int MAXN=500010;

int N,K; int p[MAXN+1],q[MAXN+1];
set<int> s;
vector<int> vec[MAXN+1];
int tr[MAXN<<2];
int deg[MAXN+1];

inline void Add(int rt,int l,int r,int k,int x){
	if(l==r){tr[rt]=min(tr[rt],x); return ;} int mid=(l+r)>>1;
	if(k<=mid) Add(rt<<1,l,mid,k,x); else Add(rt<<1|1,mid+1,r,k,x);
	tr[rt]=min(tr[rt<<1],tr[rt<<1|1]); return ;
}
inline int Query(int rt,int l,int r,int L,int R){
	if(L<=l&&R>=r) return tr[rt]; int mid=(l+r)>>1,ans=INF;
	if(L<=mid) ans=min(ans,Query(rt<<1,l,mid,L,R));
	if(R>mid) ans=min(ans,Query(rt<<1|1,mid+1,r,L,R));
	return ans;
}
inline void insert(int u,int v){
	++deg[v]; vec[u].push_back(v); return ;
}

int main(){
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	N=read(),K=read();
	for(int i=1;i<=N;i++){
		p[i]=read(); q[p[i]]=i;
	} memset(tr,127,sizeof(tr));
	for(int i=N;i>=1;i--){
		int x=Query(1,1,N,max(1,q[i]-K+1),q[i]);
		int y=Query(1,1,N,q[i],min(N,q[i]+K-1));
		if(x<=N) insert(q[i],q[x]); if(y<=N) insert(q[i],q[y]);
		Add(1,1,N,q[i],i);
	} for(int i=1;i<=N;i++) if(!deg[i]) s.insert(i); int cnt=0;
	while(s.size()){
		int k=(*s.begin()); q[++cnt]=k; s.erase(s.begin());
		for(int i=0;i<vec[k].size();i++){
			int v=vec[k][i]; if(!deg[v]) continue;
			deg[v]--; if(!deg[v]) s.insert(v);
		}
	} for(int i=1;i<=N;i++) p[q[i]]=i;
	for(int i=1;i<=N;i++) printf("%d\n",p[i]);
	return 0;
}
