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 n,k;
int a[533333],b[533333];
struct node
{
	int val;
	node *next;
}*head[533333],*tail[533333],pool[533333];
set<int> bl;
int main()
{
	ios_base::sync_with_stdio(false);
	cin>>n>>k;
	for(int i=1;i<=n;i++)
	{
		cin>>a[i];
		b[a[i]]=i;
	}
	head[0]=tail[0]=&pool[0];
	for(int i=1;i<=n;i++)
	{
		head[i]=tail[i]=&pool[i];
		head[i]->val=b[i];
	}
	bl.insert(0);
	for(int i=1;i<=k;i++)bl.insert(a[i]);
	for(int i=1;i<=n;i++)
	{
		bl.erase(a[i]);
		auto pos=*--bl.lower_bound(a[i]);
//		cerr<<"found "<<i<<' '<<pos<<endl;
		tail[pos]->next=head[a[i]];
		tail[pos]=tail[a[i]];
		if(i+k<=n)bl.insert(a[i+k]);
	}
	int top=0;
	for(node *u=head[0];u;u=u->next)
	{
		b[top++]=u->val;
	}
	for(int i=1;i<=n;i++)a[b[i]]=i;
	for(int i=1;i<=n;i++)cout<<a[i]<<endl;
	return 0;
}