GYM 101350 G

G. Snake Rana

time limit per test

4.0 s

memory limit per test

256 MB

input

standard input

output

standard output

Old Macdonald wants to build a new hen house for his hens. He buys a new rectangular area of size N by M. The night before he builds the hen house, snake Rana devises an evil plan to plant bombs in K distinct cells in the area to kill the hens and eat them for dinner later.

The morning of, Old Macdonald notices that each of the K cells, where snake Rana planted a bomb, have a marking on them. That won’t stop him though, all he must do is build the hen house in an area with no bombs.

Assume that rows are numbered from top to bottom, and columns are numbered from left to right. Old Macdonald now wants to know the number of ways he can choose sub-rectangles of top left coordinates (x1, y1) and bottom right coordinates (x2, y2) (x1 ≤ x2) (y1 ≤ y2) such that there are no bombs in the sub rectangle.

Input

The first line of input is T – the number of test cases.

The first line of each test case is three integers N, M, and K (1 ≤ N, M ≤ 104) (1 ≤ K ≤ 20).

The next K lines each contains distinct pair of integers x, y (1 ≤ x ≤ N) (1 ≤ y ≤ M) - where (x, y) is the coordinate of the bomb.

Output

For each test case, output a line containing a single integer - the number of sub-rectangles that don’t contain any bombs.

Example

input

Copy

3
2 2 1
2 2
6 6 2
5 2
2 5
10000 10000 1
1 1

output

Copy

5
257
2500499925000000
 

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstdlib>
 #include <cstring>
 #include <string>
 #include <deque>
 #include <set>
 #include <queue>
 using namespace std;
 #define  ll long long
 #define  N 10009
 #define  gep(i,a,b)   for(int  i=a;i<=b;i++)
 #define  gepp(i,a,b)  for(int  i=a;i>=b;i--)
 #define  gep1(i,a,b)  for(ll i=a;i<=b;i++)
 #define  gepp1(i,a,b) for(ll i=a;i>=b;i--)
 #define  mem(a,b)  memset(a,b,sizeof(a))
 #define  P  pair<int,int>
 #define  inf 10000009
 struct Node{
     ll x,y;
 }node[];
 int  main()
 {
     int t,k;
     ll n,m;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%lld%lld%d",&n,&m,&k);//开始时k %lld ，t直接变为0了
         ll ans=n*(n+)/*m*(m+)/;
         //一共ans个，要在减去含有炸弹的，容斥定理
         //含有炸弹的=仅含有一个的-含有两个的+含有三个的-……
         /*
 要计算几个集合并集的大小，我们要先将所有单个集合的大小
 计算出来，然后减去所有两个集合相交的部分，
 再加回所有三个集合相交的部分，
 再减去所有四个集合相交的部分，依此类推，
 一直计算到所有集合相交的部分。
 */
         gep(i,,k-){
             scanf("%lld %lld",&node[i].x,&node[i].y);
         }
         for(int i=;i<(<<k);i++)//二进制枚举所有的情况
         {
             ll m1=inf,m2=inf,m3=-,m4=-;
             int cnt=;
             gep(j,,k-){
                 if(i>>j&){//i的j位是1吗
                     m1=min(m1,node[j].x);
                     m2=min(m2,node[j].y);
                     m3=max(m3,node[j].x);
                     m4=max(m4,node[j].y);
                     cnt++;//有几个炸弹
                 }
             }
             //m1 :横坐标的最小值 m2 :纵坐标的最小值
             //m3  :横坐标的最大值  m4 :纵坐标的最大值
             //一个矩形由左上角 和右下角 决定
             //m1*m2  左上的位置数  ,(n-m3+1)*(m-m4+1):右下的位置数
             ll ret=m1*m2*(n-m3+)*(m-m4+);
             if(cnt&) ans-=ret;
             else ans+=ret;
         }
         printf("%lld\n",ans);
     }
     return ;
 }