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<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
using namespace std;
const int maxn=500010;
vector<int>g[maxn];
int n,k;
void read(int& x){
	x=0;
	char c=getchar();
	while(c<'0' || c>'9') c=getchar();
	while(c>='0' && c<='9'){
		x=x*10+c-'0';
		c=getchar();
	}
}
int p[maxn],P[maxn];
int ans[maxn];
struct Segment_Tree{
	int Max[maxn<<2];
	void modify(int o,int l,int r,int c,int pos){
		if(l==r){
			Max[o]=pos;return;
		}
		int mid=(l+r)>>1;
		if(c<=mid) modify(o<<1,l,mid,c,pos);
		else modify(o<<1|1,mid+1,r,c,pos);
		Max[o]=max(Max[o<<1],Max[o<<1|1]);
	}
	int query(int o,int l,int r,int ql,int qr){
		if(ql<=l && qr>=r) return Max[o];
		int mid=(l+r)>>1;
		int ret=0;
		if(ql<=mid) ret=query(o<<1,l,mid,ql,qr);
		if(qr>mid) ret=max(ret,query(o<<1|1,mid+1,r,ql,qr));
		return ret;
	}
}T;
int cnt=-1;
void dfs(int x){
	ans[++cnt]=x;
	sort(g[x].begin(),g[x].end());
	for(int i=0;i<g[x].size();i++) dfs(g[x][i]);
}
int main(){
	read(n);read(k);
	for(int i=1;i<=n;i++){
		read(p[i]);
		P[p[i]]=i;
	}
	P[n+1]=n+1;
	T.modify(1,1,n,P[1],1);
	for(int i=1;i<=n;i++){
		int j=T.query(1,1,n,P[i]+1,min(n,P[i]+k-1));
		if(!j) j=n+1;
		g[P[j]].push_back(P[i]);
		//printf("%d %d\n",P[j],P[i]);
		T.modify(1,1,n,P[i],i);
	}
	dfs(n+1);
	for(int i=1;i<=n;i++) p[ans[i]]=i;
	for(int i=1;i<=n;i++) printf("%d\n",p[i]);
	return 0;
}