1 #include <algorithm>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cstdio>
 5 #include <queue>
 6 using namespace std;
 7 const int maxn=200010;
 8 int cmpNo,K;
 9 struct Node{
10     int x[10],l,r,id;
11     bool operator <(const Node &b)const{
12         return x[cmpNo]<b.x[cmpNo];
13     }
14 };
15 
16 long long Dis(const Node &a,const Node &b){
17     long long ret=0;
18     for(int i=0;i<K;i++)
19         ret+=(a.x[i]-b.x[i])*(a.x[i]-b.x[i]);
20     return ret;    
21 }
22 
23 Node p[maxn];
24 
25 int Build(int l,int r,int d){
26     if(l>r)return 0;
27     cmpNo=d;
28     int mid=l+r>>1;
29     nth_element(p+l,p+mid,p+r+1);
30     p[mid].l=Build(l,mid-1,(d+1)%K);
31     p[mid].r=Build(mid+1,r,(d+1)%K);
32     return mid;
33 }
34 
35 priority_queue<pair<long long,int> >q;
36 void Kth(int l,int r,Node tar,int k,int d){
37     if(l>r)return;
38     int mid=l+r>>1;
39     pair<long long,int>v=make_pair(Dis(p[mid],tar),p[mid].id);
40     if(q.size()==k&&v<q.top())q.pop();
41     if(q.size()<k)q.push(v);
42     long long t=tar.x[d]-p[mid].x[d]; 
43     if(t<=0){
44         Kth(l,mid-1,tar,k,(d+1)%K);
45         if(q.top().first>t*t)
46             Kth(mid+1,r,tar,k,(d+1)%K);
47     }
48     else{
49         Kth(mid+1,r,tar,k,(d+1)%K);
50         if(q.top().first>t*t)
51             Kth(l,mid-1,tar,k,(d+1)%K);
52     }    
53 }
54 int k,ans[20];
55 Node a[maxn];
56 int main(){
57     int n;
58     while(scanf("%d%d",&n,&K)!=EOF){
59         for(int id=1;id<=n;id++){
60             for(int i=0;i<K;i++)
61                 scanf("%d",&p[id].x[i]);
62             p[id].id=id;
63             a[id]=p[id];    
64         }
65         Build(1,n,0);
66         int Q;
67         scanf("%d",&Q);
68         Node tar;
69         while(Q--){
70             for(int i=0;i<K;i++)
71                 scanf("%d",&tar.x[i]);
72             scanf("%d",&k);
73             Kth(1,n,tar,k,0);
74             printf("the closest %d points are:\n",k);
75             int tot=0;
76             while(!q.empty()){
77                 int id=(q.top()).second;q.pop();
78                 ans[tot++]=id;
79             }
80             for(int i=tot-1;i>=0;i--)
81                 for(int j=0;j<K;j++)
82                     printf("%d%c",a[ans[i]].x[j],j==K-1?‘\n‘:‘ ‘);        
83         }
84     }
85     return 0;
86 }