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<set>
#include<vector>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>

using namespace std;

inline int read()
{
	int x=0;char ch=getchar();
	while(ch<'0' || '9'<ch)ch=getchar();
	while('0'<=ch && ch<='9')x=x*10+(ch^48),ch=getchar();
	return x;
}

const int N=500009;

int n,k,ind[N];
int p[N],q[N];
set<int> f,b;
vector<int> g[N];

typedef set<int>::iterator it;

int main()
{
	n=read();
	k=read();

	for(int i=1;i<=n;i++)
		q[p[i]=read()]=i;
	for(int i=1;i<k;i++)
		f.insert(p[i]);
	for(int i=1;i<=n;i++)
	{
		if(i+k-1<=n)
			f.insert(p[i+k-1]);
		if(i)
			f.erase(p[i-1]);

		b.insert(p[i]);
		if(i>k)
			b.erase(p[i-k]);

		it cur=f.upper_bound(p[i]);
		if(cur!=f.end())
			g[q[*cur]].push_back(i),ind[i]++;
		cur=b.upper_bound(p[i]);
		if(cur!=b.end())
			g[q[*cur]].push_back(i),ind[i]++;
	}

	f.clear();
	for(int i=1;i<=n;i++)
		if(!ind[i])
			f.insert(-i);

	int top=0;
	while(f.size())
	{
		int u=-(*f.begin());
		p[++top]=u;
		f.erase(f.begin());
		for(int i=0,ed=g[u].size();i<ed;i++)
			if(--ind[g[u][i]]==0)
				f.insert(-g[u][i]);
	}

	for(int i=1;i<=n;i++)
		q[p[i]]=n-i+1;
	for(int i=1;i<=n;i++)
		printf("%d\n",q[i]);
	return 0;
}
