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#include <bits/stdc++.h>
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using namespace std;
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const int N = 110;
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int a[N][N];
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int main() {
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int n, m, k; //n*n的方阵，m 个火把和 k 个萤石
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cin >> n >> m >> k;
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//火把
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for (int i = 1; i <= m; i++) {
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//火把位置
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int x, y;
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cin >> x >> y;
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//火把的办量是有限的，只能向上两行，向下两行，shit~
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if (x - 1 >= 1)a[x - 1][y] = 1;
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if (x - 2 >= 1)a[x - 2][y] = 1;
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if (x + 1 <= n)a[x + 1][y] = 1;
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if (x + 2 <= n)a[x + 2][y] = 1;
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//火把的办量是有限的，只能向左两列，向右两列，shit~
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if (y - 1 >= 1)a[x][y - 1] = 1;
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if (y - 2 >= 1)a[x][y - 2] = 1;
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if (y + 1 <= n)a[x][y + 1] = 1;
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if (y + 2 <= n)a[x][y + 2] = 1;
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//斜着的还有4个方向，距离只能为1
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if (x - 1 >= 1 && y - 1 >= 1)a[x - 1][y - 1] = 1;//左上
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if (x - 1 >= 1 && y + 1 <= n)a[x - 1][y + 1] = 1;//左下
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if (x + 1 <= n && y - 1 >= 1)a[x + 1][y - 1] = 1;//右上
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if (x + 1 <= n && y + 1 <= n)a[x + 1][y + 1] = 1;//右下
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//火把的位置
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a[x][y] = 1;
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}
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//萤石
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for (int i = 1; i <= k; i++) {
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int x, y;
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cin >> x >> y;
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//萤石这个玩意比火把厉害多了啊！
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for (int j = x - 2; j <= x + 2; j++)
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for (k = y - 2; k <= y + 2; k++)
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if (j >= 1 && k >= 1 && j <= n && k <= n)
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a[j][k] = 1;
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//因为这两层循环，把a[x][y]已经带上了，就没有必要再次a[x][y]=1了
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}
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//统计没有出现过的位置，就是怪物可能出现的位置
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int cnt = 0;
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= n; ++j)
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if (!a[i][j])cnt++;
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//输出大吉
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printf("%d", cnt);
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return 0;
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}
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