/*

     BFS：先把1的入队，每个1和它相邻的组合后看看能不能使0变1，若有则添加入队，change函数返回改变了多少个0

     注意：结果还要加上原来占领的

 */

 #include <cstdio>

 #include <algorithm>

 #include <cstring>

 #include <cmath>

 #include <vector>

 #include <queue>

 #include <set>

 using namespace std;

 const int MAXN = 5e2 + ;

 const int INF = 0x3f3f3f3f;

 int a[MAXN][MAXN];

 int dx[] = {-, -, , };

 int dy[] = {-, , -, };

 int n, m;

 queue<pair<int, int> > Q;

 int change(int x, int y)

 {

     int cnt = ;

     for (int i=; i<; ++i)

     {

         int tx = x + dx[i];    int ty = y + dy[i];

         if (tx <  || tx > n)    continue;

         if (ty <  || ty > m)    continue;

         if (a[tx][ty] == )

         {

              if (i ==  || i == )

              {

                 if (a[tx][ty+] == )    {a[tx][ty+] = ;    ++cnt;    Q.push (make_pair (tx, ty+));}

                 if (a[x][y-] == )    {a[x][y-] = ;    ++cnt;    Q.push (make_pair (x, y-));}

              }

              else if (i ==  || i == )

              {

                 if (a[tx][ty-] == )    {a[tx][ty-] = ;    ++cnt;    Q.push (make_pair (tx, ty-));}

                 if (a[x][y+] == )    {a[x][y+] = ;    ++cnt;    Q.push (make_pair (x, y+));}

              }

         }

     }

     return cnt;

 }

 int BFS(void)

 {

     int ans = ;

     while (!Q.empty ())

     {

         int x = Q.front ().first;    int y = Q.front ().second;    Q.pop ();

         ans += change (x, y);

     }

     return ans;

 }

 int main(void)        //2015百度之星初赛2 HDOJ 5254 棋盘占领

 {

     int t, cas = ;    scanf ("%d", &t);

     while (t--)

     {

         while (!Q.empty ())    Q.pop ();

         scanf ("%d%d", &n, &m);

         memset (a, , sizeof (a));

         int g;    scanf ("%d", &g);

         int cnt = ;

         while (g--)

         {

             int x, y;    scanf ("%d%d", &x, &y);

             if (a[x][y] == )

             {

                 cnt++;    a[x][y] = ;    Q.push (make_pair (x, y));

             }

         }

         printf ("Case #%d:\n", ++cas);

         printf ("%d\n", BFS () + cnt);

     }

     return ;

 }

 /*

 4

 2 2

 2

 1 1

 2 2

 3 3

 3

 1 1

 2 3

 3 2

 2 4

 5

 1 1

 1 1

 1 2

 1 3

 1 4

 2 4

 2

 1 1

 2 4

 */