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 <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <set>
using namespace std;
const int N=500010;
int tt;
set <int> w;
int p[N],q[N],head[N],to[N*2],nxt[N*2],in[N];
priority_queue <int> Q;
inline int gi() {
    int x=0,o=1;
    char ch=getchar();
    while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
    if(ch=='-') ch=getchar(),o=-1;
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*o;
}
inline void lnk(int x,int y) {
    to[++tt]=y,nxt[tt]=head[x],head[x]=tt,++in[y];
}
int main() {
    int n,k;
    cin>>n>>k;
    set <int> :: iterator j;
    for(int i=1;i<=n;i++) q[p[i]=gi()]=i;
    for(int i=1;i<=n;i++) {
	if(i>k) w.erase(p[i-k]);
	j=w.insert(p[i]).first;
	if(++j!=w.end()) lnk(q[*j],i);
    }
    for(int i=n;i;i--) {
	if(i+k<=n) w.erase(p[i+k]);
	j=w.insert(p[i]).first;
	if(++j!=w.end()) lnk(q[*j],i);
    }
    for(int i=1;i<=n;i++)
	if(!in[i]) Q.push(i);
    for(int t=n;t;t--) {
	int x=Q.top();Q.pop();
	p[x]=t;
	for(int i=head[x];i;i=nxt[i])
	    if(!--in[to[i]]) Q.push(to[i]);
    }
    for(int i=1;i<=n;i++) printf("%d\n",p[i]);
    return 0;
}