1 #include<iostream>
 2 #include<stdio.h>
 3 #include<string.h>
 4 #include<math.h>
 5 #include<algorithm>
 6 using namespace std;
 7 const int N=40;
 8 struct point
 9 {
10     double x,y;
11     double angel;
12     int id;
13 } p[N],stack[N];
14 int top,n;
15 
16 double dis(point a,point b)//求距离
17 {
18     return sqrt ((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
19 }
20 
21 bool mult(point p1,point p2,point p0)//叉乘
22 {
23     return (p1.x-p0.x)*(p2.y-p0.y) >= (p2.x-p0.x)*(p1.y-p0.y);
24 }
25 
26 bool cmp(point a,point b)
27 {
28     if(a.angel == b.angel)
29     {
30         if (a.x == b.x)
31             return a.y > b.y;
32         return a.x > b.x;
33     }
34     return a.angel < b.angel;
35 }
36 
37 void graham()
38 {
39 //p为点集，n为点的个数，stack为凸包点集，top为凸包个数
40     int i,k=0;
41     point temp;
42     for(i=0; i<n; i++)
43         if(p[i].y<p[k].y||((p[i].y==p[k].y)&&(p[i].x<p[k].x)))
44             k=i;
45     temp=p[0];
46     p[0]=p[k];
47     p[k]=temp;
48     for(i=1; i<n; i++)
49         p[i].angel=atan2(p[i].y-p[0].y,p[i].x-p[0].x);
50     sort(p+1,p+n,cmp);
51     stack[0]=p[0];
52     stack[1]=p[1];
53     stack[2]=p[2];
54     top=3;
55     for(i=3; i<n; i++)
56     {
57         while(top > 2 && mult(stack[top-2],stack[top-1],p[i])<=0)
58             top--;
59         stack[top++]=p[i];
60     }
61 }
62 int main()
63 {
64     int i,j,t,pp,q,cnt,ans;
65     int flag[500];
66     scanf("%d",&t);
67     while(t--)
68     {
69         cnt=0;
70         scanf("%d%d%d",&pp,&q,&n);
71         for(i=0; i<n; i++)
72         {
73             scanf("%lf%lf",&p[i].x,&p[i].y);
74             p[i].id=i;
75         }
76         graham();
77         memset(flag,0,sizeof(flag));
78         for(i=0; i<top; i++)
79         {
80             flag[stack[i].id]=1;
81         }
82         flag[n]=flag[0];
83         for(i=0; i<n; i++)
84         {
85             if(flag[i]==1&&flag[i+1]==0)
86                 cnt++;
87         }
88         ans=q*(top-cnt)-pp*cnt;
89         if(ans<=0)
90             printf("0\n");
91         else
92             printf("%d\n",ans);
93     }
94     return 0;
95 }