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 <stdio.h>
#include <stdlib.h>
#include <algorithm>
#include <string.h>
#include <set>
using namespace std;
#define SZ 666666
//只有单调栈内的才有ls
int n,k,p[SZ],s[SZ],ls[SZ],nx[SZ],rv[SZ],op[SZ];
typedef pair<int,int> pii;
set<pii> si;
#define FO(x) {freopen(#x".in","r",stdin);freopen(#x".out","w",stdout);}
int main()
{
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;++i)
		scanf("%d",p+i),s[p[i]]=i,ls[i]=i;
	for(int i=1;i<=n;++i)
	{
		int g=s[i]+k;
		set<pii>::iterator p=si.lower_bound(pii(g,0));
		if(p!=si.begin())
		{
			--p;
			while(1)
			{
				if(p->first<=s[i]) break;
				int x=p->second;
				nx[x]=ls[i]; ls[i]=ls[x];
				si.erase(p++);
				if(p==si.begin()) break;
				--p;
			}
		}
		si.insert(pii(s[i],i));
	}
	int on=0;
	for(set<pii>::iterator i=si.begin();
	i!=si.end();++i)
	{
		int r=i->second;
		for(int p=ls[r];p;p=nx[p]) op[++on]=s[p];
	}
	for(int i=1;i<=on;++i) rv[op[i]]=i;
	for(int i=1;i<=n;++i) printf("%d\n",rv[i]);
}
/*
定义临界态：s[i]<s[i+1]+K。
猜想：一个排列只能到一个临界态
*/