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<set>
#include<queue>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=5e6+10;
int a[maxn],b[maxn],c[maxn],deg[maxn],to[maxn],beg[maxn],nex[maxn],ans[maxn];
int e;
set<int> st1,st2;
priority_queue<int> q;
inline int read(){
	int x=0,flag=1;
	char ch=getchar();
	while(!isdigit(ch) && ch!='-')ch=getchar();
	if(ch=='-')flag=-1,ch=getchar();
	while(isdigit(ch))x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
	return x*flag;
}
inline void add(int x,int y){
	to[++e]=y;
	nex[e]=beg[x];
	beg[x]=e;
}
int main(){
	int i,j,k,m,n;
	n=read();k=read();
	for(i=1;i<=n;i++){
		a[i]=read();
		b[a[i]]=i;
	}
	for(i=1;i<k;i++)st1.insert(a[i]);
	for(i=1;i<=n;i++){
		if(i+k-1<=n)st1.insert(a[i+k-1]);
		st1.erase(a[i-1]);
		st2.insert(a[i]);
		if(i>k)st2.erase(a[i-k]);
		set<int>::iterator cur=st1.upper_bound(a[i]);
		if(cur!=st1.end()){
			add(b[*cur],i);
			deg[i]++;
		}
		cur=st2.upper_bound(a[i]);
		if(cur!=st2.end()){
			add(b[*cur],i);
			deg[i]++;
		}
	}
	for(i=1;i<=n;i++)
		if(!deg[i])q.push(i);
	int cnt=0;
	while(!q.empty()){
		int x=q.top();
		q.pop();
		ans[++cnt]=x;
		for(i=beg[x];i;i=nex[i]){
			deg[to[i]]--;
			if(deg[to[i]]==0)q.push(to[i]);
		}
	}
	for(i=1;i<=n;i++)c[ans[i]]=n-i+1;
	for(i=1;i<=n;i++)printf("%d\n",c[i]);
	return 0;
}