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<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<set>
#include<queue>
#define next Next
#define begin Begin
using namespace std;
typedef long long ll;
const int maxn=500000+10;
inline int read(){
	int x;
	char c;
	int f=1;
	while((c=getchar())!='-' && (c<'0' || c>'9'));
	if(c=='-') c=getchar(),f=-1;
	x=c^'0';
	while((c=getchar())>='0' && c<='9') x=(x<<1)+(x<<3)+(c^'0');
	return x*f;
}
priority_queue<int> q;
inline ll readll(){
	ll x;
	char c;
	ll f=1;
	while((c=getchar())!='-' && (c<'0' || c>'9'));
	if(c=='-') c=getchar(),f=-1;
	x=c^'0';
	while((c=getchar())>='0' && c<='9') x=(x<<1ll)+(x<<3ll)+(c^'0');
	return x*f;
}
int begin[maxn],next[maxn<<1],to[maxn<<1],e;
void add(int x,int y){
	to[++e]=y;
	next[e]=begin[x];
	begin[x]=e;
}
int d[maxn];
int p[maxn],a[maxn];
set<int> st1,st2;
int main(){
	int n=read(),k=read();
	for(int i=1;i<=n;i++){
		a[i]=read();
		p[a[i]]=i;
	}
	for(int i=1;i<k-1;i++) st1.insert(a[i]);
	for(int 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 it=st1.upper_bound(a[i]);
		if(it!=st1.end()) add(p[*it],i),d[i]++;
		it=st2.upper_bound(a[i]);
		if(it!=st2.end()) add(p[*it],i),d[i]++;
	}
	for(int i=1;i<=n;i++)
		if(!d[i])
			q.push(i);
	int ans=0;
	while(!q.empty()){
		int u=q.top();
		q.pop();
		a[u]=n-ans;
		ans++;
		for(int i=begin[u];i;i=next[i]){
			d[to[i]]--;
			if(!d[to[i]]) q.push(to[i]);
		}
	}
	for(int i=1;i<=n;i++)
		printf("%d\n",a[i]);
	return 0;
}