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<set>
#include<vector>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,k,ind[500003];
int p[500003],q[500003];
set<int> f,b;
vector<int> g[500003];
typedef set<int>::iterator it;
int main(){
	cin>>n>>k;
    for(int i=1;i<=n;i++){
    	cin>>p[i];
    	q[p[i]]=i;
	}
    for(int i=1;i<k;i++){
    	f.insert(p[i]);
	}
    for(int i=1;i<=n;i++){
        if(i+k-1<=n){
        	f.insert(p[i+k-1]);
		}
        if(i){
        	f.erase(p[i-1]);
		}
        b.insert(p[i]);
        if(i>k){
        	b.erase(p[i-k]);
		}
        it cur=f.upper_bound(p[i]);
        if(cur!=f.end()){
        	g[q[*cur]].push_back(i),ind[i]++;
		}
        cur=b.upper_bound(p[i]);
        if(cur!=b.end()){
        	g[q[*cur]].push_back(i),ind[i]++;
		}
    }
	f.clear();
    for(int i=1;i<=n;i++){
    	if(!ind[i]){
			f.insert(-i);
		}
	}
    int top=0;
	while(f.size()){
		int u=-(*f.begin());
		p[++top]=u;
		f.erase(f.begin());
		for(int i=0,ed=g[u].size();i<ed;i++){
			if(--ind[g[u][i]]==0){
				f.insert(-g[u][i]);
			}
		}
    }
	for(int i=1;i<=n;i++){
		q[p[i]]=n-i+1;
	}
    for(int i=1;i<=n;i++){
    	cout<<q[i]<<endl;
	}
    return 0;
}