1 //凸包稳定性判断：每条边上是否至少有三点
2 // POJ 1228
3
4 #include <iostream>
5 #include <cstdio>
6 #include <cstdlib>
7 #include <algorithm>
8 #include <vector>
9 #include <math.h>
10 using namespace std;
11 #define LL long long
12 typedef pair<int,int> pii;
13 const double inf = 0x3f3f3f3f;
14 const LL MOD =100000000LL;
15 const int N =110;
16 #define clc(a,b) memset(a,b,sizeof(a))
17 const double eps = 1e-8;
18 void fre() {freopen("in.txt","r",stdin);}
19 void freout() {freopen("out.txt","w",stdout);}
20 inline int read() {int x=0,f=1;char ch=getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch=getchar();}while(ch>='0'&&ch<='9') {x=x*10+ch-'0';ch=getchar();}return x*f;}
21
22 int sgn(double x){
23 if(fabs(x) < eps)return 0;
24 if(x < 0)return -1;
25 else return 1;
26 }
27
28 struct Point{
29 double x,y;
30 Point(){}
31 Point(double _x,double _y){
32 x = _x;y = _y;
33 }
34 Point operator -(const Point &b)const{
35 return Point(x - b.x,y - b.y);
36 }
37 double operator ^(const Point &b)const{
38 return x*b.y - y*b.x;
39 }
40 double operator *(const Point &b)const{
41 return x*b.x + y*b.y;
42 }
43 friend bool operator<(const Point &a,const Point &b){
44 if(fabs(a.y-b.y)<eps) return a.x<b.x;
45 return a.y<b.y;
46 }
47 friend double dis2(Point a){
48 return a.x*a.x+a.y*a.y;
49 }
50 };
51
52 double dis(Point a,Point b){
53 return sqrt(dis2(a-b));
54 }
55
56 int mult(Point a,Point b,Point o){
57 return sgn((a.x-o.x)*(b.y-o.y)-(b.x-o.x)*(a.y-o.y));
58 }
59 //返回top个点
60 int graham(Point p[],int n,Point q[]){
61 int top=1;
62 sort(p,p+n);
63 if(n==0) return 0;
64 q[0]=p[0];
65 if(n==1) return 1;
66 q[1]=p[1];
67 if(n==2) return 2;
68 q[2]=p[2];
69 for(int i=2;i<n;i++){
70 while(top&&(mult(p[i],q[top],q[top-1])>0)) top--;
71 q[++top]=p[i];
72 }
73 int len=top;
74 q[++top]=p[n-2];
75 for(int i=n-3;i>=0;i--){
76 while(top!=len&&(mult(p[i],q[top],q[top-1])>0)) top--;
77 q[++top]=p[i];
78 }
79 return top;
80 }
81
82 // 判断凸包边上是否至少有三点
83 bool judge(Point p[],int m){
84 p[m]=p[0];
85 p[m+1]=p[1];
86 for(int i=1;i<m;i++){
87 if(mult(p[i-1],p[i+1],p[i])!=0&&mult(p[i],p[i+2],p[i+1])!=0)
88 return false;
89 }
90 return true;
91 }
92
93
94 Point p[1011];
95 Point ch[1011];
96 int main(){
97 // fre();
98 int T;
99 scanf("%d",&T);
100 while(T--){
101 int n;
102 scanf("%d",&n);
103 for(int i=0;i<n;i++){
104 double x,y;
105 scanf("%lf%lf",&x,&y);
106 p[i]=Point(x,y);
107 }
108 int m=graham(p,n,ch);
109 if(n<6) {
110 puts("NO");
111 continue;
112 }
113 if(judge(ch,m)) puts("YES");
114 else puts("NO");
115 }
116 return 0;
117 }
