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 <bits/stdc++.h>
using namespace std;
#define ref(i,x,y)for(int i=x;i<=y;++i)
const int N=500010;
int n,m,p[N],t[N],op,ed;set<pair<int,int> > S;
struct node{int l,r;}a[N];
int main(){
	scanf("%d%d",&n,&m);
	ref(i,1,n)scanf("%d",&p[i]);
	S.insert(make_pair(n+1,n+1));
	ref(i,1,n){
		S.insert(make_pair(p[i],i));
		if(i-m>0)S.erase(make_pair(p[i-m],i-m));
		t[i]=(*S.upper_bound(make_pair(p[i],i))).second;
	}
	ed=n+1;op=n+2;a[op].r=ed;a[ed].l=op;
	ref(i,1,n){
		a[i].r=t[i];a[i].l=a[t[i]].l;
		a[a[t[i]].l].r=i;a[t[i]].l=i;
	}
	for(int i=a[op].r,s=1;i!=ed;i=a[i].r,++s)p[i]=s;
	ref(i,1,n)printf("%d\n",p[i]);
}