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# include <bits/stdc++.h>
using namespace std ;
const int N = 110 ;
int a [ N ] [ N ] ;
int main ( ) {
int n , m , k ; //n*n的方阵，
cin > > n > > m > > k ;
//火把
for ( int i = 1 ; i < = m ; i + + ) {
//火把位置
int x , y ;
cin > > x > > y ;
//火把的办量是有限的， ， ，
if ( x - 1 > = 1 ) a [ x - 1 ] [ y ] = 1 ;
if ( x - 2 > = 1 ) a [ x - 2 ] [ y ] = 1 ;
if ( x + 1 < = n ) a [ x + 1 ] [ y ] = 1 ;
if ( x + 2 < = n ) a [ x + 2 ] [ y ] = 1 ;
//火把的办量是有限的， ， ，
if ( y - 1 > = 1 ) a [ x ] [ y - 1 ] = 1 ;
if ( y - 2 > = 1 ) a [ x ] [ y - 2 ] = 1 ;
if ( y + 1 < = n ) a [ x ] [ y + 1 ] = 1 ;
if ( y + 2 < = n ) a [ x ] [ y + 2 ] = 1 ;
//斜着的还有4个方向，
if ( x - 1 > = 1 & & y - 1 > = 1 ) a [ x - 1 ] [ y - 1 ] = 1 ; //左上
if ( x - 1 > = 1 & & y + 1 < = n ) a [ x - 1 ] [ y + 1 ] = 1 ; //左下
if ( x + 1 < = n & & y - 1 > = 1 ) a [ x + 1 ] [ y - 1 ] = 1 ; //右上
if ( x + 1 < = n & & y + 1 < = n ) a [ x + 1 ] [ y + 1 ] = 1 ; //右下
//火把的位置
a [ x ] [ y ] = 1 ;
}
//萤石
for ( int i = 1 ; i < = k ; i + + ) {
int x , y ;
cin > > x > > y ;
//萤石这个玩意比火把厉害多了啊！
for ( int j = x - 2 ; j < = x + 2 ; j + + )
for ( k = y - 2 ; k < = y + 2 ; k + + )
if ( j > = 1 & & k > = 1 & & j < = n & & k < = n )
a [ j ] [ k ] = 1 ;
//因为这两层循环， ，
}
//统计没有出现过的位置，就是怪物可能出现的位置
int cnt = 0 ;
for ( int i = 1 ; i < = n ; i + + )
for ( int j = 1 ; j < = n ; + + j )
if ( ! a [ i ] [ j ] ) cnt + + ;
//输出大吉
printf ( " %d " , cnt ) ;
return 0 ;
}
