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;
int read()
{
	char s;
	int k=0,base=1;
	while((s=getchar())!='-'&&s!=EOF&&!(s>='0'&&s<='9'));
	if(s==EOF)exit(0);
	if(s=='-')base=-1,s=getchar();
	while(s>='0'&&s<='9')
	{
		k=k*10+(s-'0');
		s=getchar();
	}
	return k*base;
}
void write(int x)
{
	if(x<0)
	{
		putchar('-');
		write(-x);
	}
	else
	{
		if(x/10)write(x/10);
		putchar(x%10+'0');
	}
}
const int maxn=500010;
const int maxm=1010010;
int n,k,T;
int rd[maxn];
int p[maxn],q[maxn];
int ans[maxn];
int id,to[maxm<<1],ne[maxm<<1],po[maxn];
inline void add(int x,int y)
{
	id++;
	to[id]=y;ne[id]=po[x];
	po[x]=id;
	rd[y]++;
}
set<int> s;
set<int>::iterator it;
priority_queue<int,vector<int>,greater<int> > Q;
int main()
{
	n=read();k=read();
	for (int i=1;i<=n;i++) q[p[i]=read()]=i;
	for (int i=1;i<=n;i++)
	{
		if (i>k) s.erase(p[i-k]);
		it=s.insert(p[i]).first;
		if (++it!=s.end()) add(i,q[*it]);
	}
	for (int i=n;i>=1;i--)
	{
		if (i+k<=n) s.erase(p[i+k]);
		it=s.insert(p[i]).first;
		if (++it!=s.end()) add(i,q[*it]);
	}
	for (int i=1;i<=n;i++)
	{
		if (rd[i]==0) Q.push(i);
	}
	for (int i=1;i<=n;i++)
	{
		T=Q.top();Q.pop();
		ans[T]=i;
		for (int j=po[T];j;j=ne[j])
		{
			rd[to[j]]--;
			if (rd[to[j]]==0) Q.push(to[j]);
		}
	}
//	printf("-----------------------\n");
	for (int i=1;i<=n;i++)
	{
		printf("%d\n",ans[i]);
	}
	return 0;
}
