[HDU

　　题意：

　　　　给出一个由n个点组成的凸包，以及凸包内的k个点，问是否能够在凸包上选择最多2k个点构造一个新的

　　凸包，使得新的凸包覆盖原来的k个点。

　　　　

　　　　要求2k个点覆盖原本的k个点，只要对原k个点构造凸包，然后选择该凸包内一点O与该凸包的顶点连一条射线，其

　　与大凸包相交的边的两端点即为要保留的点，小凸包的顶点个数<=k，所有结果最后一定存在。

　　

　　　　

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<time.h>
#include<cstdlib>
#include<cmath>
#include<list>
using namespace std;
#define MAXN 100100
#define eps 1e-7
#define For(i,a,b) for(int i=a;i<=b;i++) 
#define Fore(i,a,b) for(int i=a;i>=b;i--) 
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define mkp make_pair
#define pb push_back
#define cr clear()
#define sz size()
#define met(a,b) memset(a,b,sizeof(a))
#define iossy ios::sync_with_stdio(false)
#define fre freopen
#define pi acos(-1.0)
#define inf 1e9+9
#define Vector Point
const int Mod=1e9+7;
typedef unsigned long long ull;
typedef long long ll;
int dcmp(double x){
    if(fabs(x)<=eps) return 0;
    return x<0?-1:1;
}
double ox,oy;
struct Point{
    double x,y,id;
    Point(){}
    Point(double x,double y):x(x),y(y) {}
    bool operator < (const Point &a)const { if(x==a.x) return y<a.y; return x<a.x; }
    Point operator - (const Point &a)const{ return Point(x-a.x,y-a.y); }
    Point operator + (const Point &a)const{ return Point(x+a.x,y+a.y); }
    Point operator * (const double &a)const{ return Point(x*a,y*a); }
    Point operator / (const double &a)const{ return Point(x/a,y/a); }
    void read(){ scanf("%lf%lf",&x,&y); }
    bool operator == (const Point &a)const{ return dcmp(x-a.x)==0 && dcmp(y-a.y)==0; }
};
double Dot(Vector a,Vector b) { return a.x*b.x+a.y*b.y; }
ll dis(Vector a) { return sqrt(Dot(a,a)); }
double Cross(Vector a,Vector b) { return a.x*b.y-a.y*b.x; }
int ConvexHull(Point *p,int n,Point *ch){
    sort(p+1,p+1+n);
    int m=0;
    For(i,1,n){
        while(m>1 && Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;
        ch[m++]=p[i];
    }
    int k=m;
    Fore(i,n-1,1) {
        while(m>k && Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;
        ch[m++]=p[i];
    }
    if(n>1) m--;
    return m;
}
Point getInsect(Point p,Vector v,Point q,Vector w) {
    Vector u=p-q;
    double t=Cross(w,u)/Cross(v,w);
    return p+v*t;
}
Point p[200005],pp[200005],ch[200005],o;
bool chk(Point p1,Point p2,Point q1,Point q2) {
    double c1=Cross(p2-p1,q1-p1),c2=Cross(p2-p1,q2-p1);    
    double c3=Dot(p2-p1,q1-p1),c4=Dot(p2-p1,q2-p1);
    if(dcmp(c1)*dcmp(c2)<0) {
        Point cp=getInsect(p1,p2-p1,q1,q2-q1);
        return Dot(cp-p1,p2-p1)>0;
    }
    return 0;
}
int n,k,res;
set<int>ans;
void solve(){
    ans.clear();
    scanf("%d",&n);
    For(i,0,n-1) p[i].read();
    scanf("%d",&k);
    For(i,1,k) pp[i].read();
    int m=ConvexHull(pp,k,ch);
    o=ch[0];
    For(i,1,m-1) o=(o+ch[i])/2;
    res=0;
    For(i,0,m-1) {
        while(!chk(o,ch[i],p[res],p[(res+1)%n])) res=(res+1)%n;
        ans.insert(res);ans.insert((res+1)%n);
    }
    set<int>::iterator it;
    cout<<"Yes"<<endl;
    cout<<ans.sz<<endl;
    for(it=ans.begin();it!=ans.end();it++){
        if(it!=ans.begin()) printf(" ");
        printf("%d",(*it)+1);
    }
    cout<<endl;
}
int main(){
//    fre("in.txt","r",stdin);
    int t=1;
    cin>>t;
    For(i,1,t) solve();
    return 0;
}