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<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
typedef long long ll;
#define N 1000002
const int p=1e9+7;
struct node{int x,va;friend bool operator <(node n1,node n2){return n1.va<n2.va;}};
struct edge{int to,nxxt;}e[N<<1];
int n,a[N],b[N],head[N],cnt=1,rd[N],s[N],tot,k;
priority_queue<node>q;
inline void ins(int u,int v){e[cnt].to=v;e[cnt].nxxt=head[u];head[u]=cnt++;}
int main()
{
	scanf("%d%d",&n,&k);if(n>1000)return;
	for(int i=1;i<=n;i++)scanf("%d",&a[i]),b[a[i]]=i;
	for(int i=1,j=n;i<j;i++,j--)swap(b[i],b[j]);
	for(int i=1;i<=n;i++)for(int j=i+1;j<=n;j++)if(abs(b[i]-b[j])<k)ins(i,j),rd[j]++;
	for(int i=1;i<=n;i++)if(!rd[i])q.push((node){i,b[i]});
	while(!q.empty())
	{
		node a1=q.top();q.pop();s[++tot]=a1.va;;
		for(int i=head[a1.x];i;i=e[i].nxxt)
		{
			int t1=e[i].to;rd[t1]--;if(!rd[t1])q.push((node){t1,b[t1]});
		}
	}
	for(int i=1,j=n;i<j;i++,j--)swap(s[i],s[j]);
	for(int i=1;i<=n;i++)a[s[i]]=i;
	for(int i=1;i<=n;i++)printf("%d\n",a[i]);
}
/*
10 3
7 5 3 4 10 8 1 6 2 9 
*/