Description

Wintokk has collected a huge amount of coins at THU. One day he had all his coins fallen on to the ground. Unfortunately, WangDong came by and decided to rob Wintokk of the coins.

They agreed to distribute the coins according to the following rules:

Consider the ground as a plane. Wintokk draws a horizontal line on the plane and then WangDong draws a vertical one so that the plane is divided into 4 parts, as shown below.

Wintokk will save the coins in I and III while thZ??http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">vc2UgZml0IGluIElJIGFuZCBJViB3aWxsIGJlIHRha2VuIGF3YXkgYnkgdGhlIHJvYmJlciBXYW5nRG9uZy48L3A+CjxwPkZvciBmaXhlZCBsb2NhdGlvbnMgb2YgdGhlIGNvaW5zIG93bmVkIGJ5IFdpbnRva2ssIHRoZXkgZHJldyBzZXZlcmFsIHBhaXJzIG9mIGxpbmVzLiBGb3IgZWFjaCBwYWlyLCBXaW50b2trIHdhbnRzIHRvIGtub3cgdGhlIGRpZmZlcmVuY2UgYmV0d2VlbiB0aGUgbnVtYmVyIG9mIHRoZSBzYXZlZCBjb2lucyBhbmQgdGhhdCBvZiB0aGUgbG9zdCBjb2lucy48L3A+CjxwPkl0"s guaranteed that all the coins will lie on neither of the lines drew by that two guys.

Input

The first line contains an integer T, indicating the number of test cases. Then T blocks of test cases follow. For each case, the first line contains two integers N and M, where N is the number of coins on the ground and M indicates how many times they are going to draw the lines. The 2nd to (N+1)-th lines contain the co-ordinates of the coins and the last M lines consist of the M pairs integers (x, y) which means that the two splitting lines intersect at point (x, y).

(N,M ≤ 50000, 0 ≤x,y ≤ 500000)

Output

For each query, output a non-negative integer, the difference described above. Output a blank line between cases.

Sample Input

2
10 3
29 22
17 14
18 23
3 15
6 28
30 27
4 1
26 7
8 0
11 21
2 25
5 10
19 24
10 5
28 18
2 29
6 5
13 12
20 27
15 26
11 9
23 25
10 0
22 24
16 30
14 3
17 21
8 1
7 4

Sample Output

6
4
4

2
2
4
4
4

Source

#include
  
   
#include
   
     #include
    
      #include
     
       using namespace std; int l[500005],r[500005]; struct P { int x; int y; int id; }point[50005],coin[50005]; int cmp(P a,P b) { if(a.x!=b.x) return a.x
      
       0;i-=lowbit(i)) sum+=s[i]; return sum; } int main() { int i,t,n,m,ans[50005],y; scanf("%d",&t); while(t--) { y=-1; memset(l,0,sizeof(l)); memset(r,0,sizeof(r)); memset(ans,0,sizeof(ans)); scanf("%d%d",&n,&m); for(i=1;i<=n;i++) { scanf("%d%d",&coin[i].x,&coin[i].y); coin[i].x++;coin[i].y++; if(y
       
        =coin[op].x&&op<=n) { add(l,coin[op].y,y,1); add(r,coin[op].y,y,-1); op++; } int a=sum(r,y)-sum(r,point[i].y)+sum(l,point[i].y); int b=sum(l,y)-sum(l,point[i].y)+sum(r,point[i].y); ans[point[i].id]=abs(a-b); } for(i=1;i<=m;i++) printf("%d\n",ans[i]); printf("\n"); } return 0; }