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<cstring>
#define N 1048576
inline int min(int x,int y){return x<y?x:y;}
inline int max(int x,int y){return x<y?y:x;}
char G[3500000],*H=G;
inline int read()
{
	int x=0;
	while(*H<48||57<*H)H++;
	while(47<*H&&*H<58)x=(x<<1)+(x<<3)+(*H++^48);
	return x;
}
int S[6],T;
inline void print(int x)
{
	int y;
	do y=x/10,S[T++]=x-y*10,x=y;while(x);
	while(T)*H++=S[--T]^48;
}
int a[N],b[N],g[N],h[N],i,j,k,m,n,o;
inline void up(int u){g[u]=min(g[u<<1],g[u<<1|1]);}
inline void add(int u,int v){g[u]+=v,h[u]+=v;}
inline void down(int u){if(h[u])add(u<<1,h[u]),add(u<<1|1,h[u]),h[u]=0;}
inline void push(int u){for(j=o;j;j--)down(u>>j);}
inline void add(int u,int v,int w)
{
	for(push(u+=m-1),push(v+=m+1);u^v^1;up(u>>=1),up(v>>=1))
	{
		if(~u&1)add(u^1,w);
		if(v&1)add(v^1,w);
	}
	while(u>>=1)up(u);
}
inline int query()
{
	int u=1;
	for(;u<m;u=u<<1|(g[u<<1]>0))down(u);
	return u^m;
}
inline void add(int u){while(u<=n)h[u]++,u+=u&-u;}
inline void del(int u){while(u<=n)h[u]--,u+=u&-u;}
inline int sum(int u)
{
	int x=0;
	while(u)x+=h[u],u^=u&-u;
	return x;
}
int main()
{
	fread(G,1,3500000,stdin),n=read(),k=read();
	for(i=1;i<=n;i++)b[a[i]=read()]=i;
	for(m=1;m<n+2;m<<=1)o++;
	for(i=1;i<=n;i++)g[m|b[i]]=sum(min(n,b[i]+k-1))-sum(max(0,b[i]-k)),add(b[i]);
	for(i=g[m]=n+1,memset(h+1,0,n<<2);i<m;i++)g[m|i]=n+1;
	for(i=m,H=G;--i;)up(i);
	for(i=1;i<=n;i++)a[b[i]=query()]=i,add(b[i],b[i],n+1),add(max(1,b[i]-k+1),min(n,b[i]+k-1),-1);
	for(i=1;i<=n;*H++='\n',i++)print(a[i]);
	return 0&fwrite(G,1,H-G,stdout);
}