考虑k比较小，想到容斥原理，枚举，容斥求和。
//http://www.cnblogs.com/IMGavin/
#include <iostream>
#include <stdio.h>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <vector>
#include <map>
#include <stack>
#include <set>
#include <bitset>
#include <algorithm>
using namespace std;
typedef long long LL;
#define gets(A) fgets(A, 1e8, stdin)
const int INF = 0x3F3F3F3F, N = 1008, MOD = 1003;
const double EPS = 1e-6;
int x[N], y[N];
int tot;
int r[N], c[N], n, m, k;
LL ans;
void dfs(int d, int cnt, int x1, int y1){
if(d == tot){
if(cnt & 1){
ans -= x1 * y1;
}else{
ans += x1 * y1;
}
return ;
}
dfs(d + 1, cnt, x1, y1);
dfs(d + 1, cnt + 1, min(x1, x[d]), min(y1, y[d]));
}
int main(){
while(cin >> n >> m >> k){
for(int i = 0; i < k; i++){
scanf("%d %d", &r[i], &c[i]);
}
ans = 0;
for(int i = 1; i <= n; i++){
for(int j = 1; j <= m; j++){
tot = 0;
bool ok = 1;
for(int l = 0; l < k; l++){
if(i == r[l] && j == c[l]){
ok = 0;
break;
}
if(i >= r[l] && j >= c[l]){
x[tot] = r[l];
y[tot] = c[l];
tot++;
}
}
if(ok){
dfs(0, 0, i, j);
}
}
}
cout<<ans<<endl;
}
return 0;
}