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<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<set>
using namespace std;
const int MAXN=500000+10;
int n,k,A[MAXN],P[MAXN],Indegree[MAXN],to[MAXN<<1],nex[MAXN<<1],beg[MAXN],Ans[MAXN],cnt,e;
priority_queue<int> q;
set<int> tmp1,tmp2;
inline void read(int &x)
{
	int data=0,w=1;
	char ch=0;
	while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
	if(ch=='-')w=-1,ch=getchar();
	while(ch>='0'&&ch<='9')data=(data<<3)+(data<<1)+(ch^'0'),ch=getchar();
	x=data*w;
}
inline void insert(int x,int y)
{
	to[++e]=y;
	nex[e]=beg[x];
	beg[x]=e;
}
inline void toposort()
{
	while(!q.empty())
	{
		int x=q.top();
		q.pop();
		Ans[++cnt]=x;
		for(register int i=beg[x];i;i=nex[i])
		{
			Indegree[to[i]]--;
			if(!Indegree[to[i]])q.push(to[i]);
		}
	}
}
int main()
{
	read(n);read(k);
	for(register int i=1;i<=n;++i)
	{
		read(P[i]);
		A[P[i]]=i;
	}
	for(register int i=1;i<k-1;i++)tmp1.insert(P[i]);
	for(register int i=1;i<=n;i++)
	{
		if(i+k-1<=n)tmp1.insert(P[i+k-1]);
		tmp1.erase(P[i-1]);
		tmp2.insert(P[i]);
		if(i>k)tmp2.erase(P[i-k]);
		set<int>::iterator cur=tmp1.upper_bound(P[i]);
		if(cur!=tmp1.end())insert(A[*cur],i),Indegree[i]++;
		cur=tmp2.upper_bound(P[i]);
		if(cur!=tmp2.end())insert(A[*cur],i),Indegree[i]++;
	}
	for(register int i=1;i<=n;++i)
		if(!Indegree[i])q.push(i);
	toposort();
	for(register int i=1;i<=n;++i)A[Ans[i]]=n-i+1;
	for(register int i=1;i<=n;++i)printf("%d\n",A[i]);
	return 0;
}
