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#include <bits/stdc++.h>
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using namespace std;
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typedef pair<int, int> PII;
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#define x first
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#define y second
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const int N = 110;
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int n, m, k;
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PII match[N][N];
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int g[N][N], st[N][N];
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int dx[] = {-2, -1, 1, 2, 2, 1, -1, -2};
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int dy[] = {1, 2, 2, 1, -1, -2, -2, -1};
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int find(int x, int y) {
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for (int i = 0; i < 8; i++) {
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int tx = x + dx[i], ty = y + dy[i];
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if (tx < 1 || tx > n || ty < 1 || ty > m) continue;
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if (g[tx][ty] || st[tx][ty]) continue;
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// 男[x,y] 找到女[tx,ty]
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st[tx][ty] = 1;
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// t为女[tx,ty]现在匹配的对象
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PII t = match[tx][ty];
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// 如果女[tx,ty]没有匹配对象,或者,现配t可以去找其他妹子,那就把[tx,ty]给[x,y]
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if (t.x == 0 || find(t.x, t.y)) {
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match[tx][ty] = {x, y};
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return 1;
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}
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}
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return 0;
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}
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int main() {
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scanf("%d %d %d", &n, &m, &k);
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for (int i = 1; i <= k; i++) {
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int x, y;
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scanf("%d %d", &x, &y);
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g[x][y] = 1; // 不可以放置
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}
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int res = 0;
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= m; j++) {
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//(i,j)位置不可以放 只看i+j是奇数的点（偶数的点是一样的）
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if ((i + j) % 2 && !g[i][j]) {
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memset(st, 0, sizeof st);
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if (find(i, j)) res++; // 开始跑匈牙利算法
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}
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}
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// 最大独立集 n-无法放的点-最大匹配数
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printf("%d\n", n * m - k - res);
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return 0;
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}
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