2 5 10
2 2
1 1
1 4
2 1
1 2
1 5
2 4
2 5
1 3
2 3
 2 2 2
1 1
2 2
 3 3 9
1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3
 7 5 35
5 3
3 5
1 1
2 5
6 4
7 2
5 1
3 4
6 5
7 3
4 1
2 2
4 2
2 4
4 3
3 3
6 1
1 2
4 5
1 3
2 1
5 5
7 5
7 1
6 2
1 4
1 5
3 2
4 4
5 2
2 3
6 3
7 4
5 4
3 1
 1 5 4
1 1
1 2
1 4
1 5
 1 1 1
1 1
 5 7 21
4 4
4 5
5 5
5 6
5 7
5 1
5 2
4 2
3 2
3 1
3 7
3 6
2 6
1 6
1 7
1 1
1 2
1 3
1 4
2 4
3 4
 2 1 1
1 1
 2 2 4
2 1
1 1
1 2
2 2
 3000 3000 1
2222 738
 5 7 35
1 2
2 3
5 1
5 7
4 4
5 3
3 1
2 6
4 5
2 7
4 1
2 1
2 2
1 7
5 6
3 6
4 6
1 6
1 1
3 3
3 4
4 2
2 5
5 5
3 5
2 4
1 5
1 3
4 3
5 4
1 4
3 7
4 7
3 2
5 2
 5 4 1
4 1
 3 4 12
2 3
3 2
2 2
2 1
1 2
3 1
1 4
1 3
3 3
2 4
3 4
1 1
 3 3 3
1 1
2 2
3 3

8
 1
 6
 26
 4
 0
 20
 0
 2
 1
 28
 1
 9
 3

#include <bits/stdc++.h>
using namespace std;
const int inf = ~0U >> 1;
const long long INF = ~0ULL >> 1;
template <class T>
inline void read(T &n) {
  char c;
  for (c = getchar(); !(c >= '0' && c <= '9'); c = getchar())
    ;
  n = c - '0';
  for (c = getchar(); c >= '0' && c <= '9'; c = getchar()) n = n * 10 + c - '0';
}
int Pow(int base, int n, int mo) {
  if (n == 0) return 1;
  if (n == 1) return base % mo;
  int tmp = Pow(base, n >> 1, mo);
  tmp = (long long)tmp * tmp % mo;
  if (n & 1) tmp = (long long)tmp * base % mo;
  return tmp;
}
int n, m, ans, K;
int id[3100][6100];
int f[3100 * 6100], tmp[3100 * 6100], buff[3100 * 6100], top, v[3100][6100];
const int dx[8] = {1, 1, 1, -1, -1, -1, 0, 0};
const int dy[8] = {1, -1, 0, 1, -1, 0, 1, -1};
int find(int x) {
  if (f[x] == x) return f[x];
  int t = find(f[x]);
  buff[++top] = x;
  tmp[top] = f[x];
  f[x] = t;
  return t;
}
void Union(int x, int y) {
  x = find(x);
  y = find(y);
  buff[++top] = x;
  tmp[top] = f[x];
  f[x] = y;
}
void add(int x, int y) {
  for (int k = (0); k <= (7); ++k) {
    int nx = x + dx[k];
    int ny = y + dy[k];
    if (nx > 0 && ny > 0 && nx <= n && ny <= 2 * m) {
      if (v[nx][ny]) Union(id[x][y], id[nx][ny]);
    }
  }
  v[x][y] = 1;
}
int main() {
  scanf("%d%d%d", &n, &m, &K);
  for (int i = (1); i <= (n); ++i)
    for (int j = (1); j <= (2 * m); ++j)
      id[i][j] = (i - 1) * 2 * m + j, f[id[i][j]] = id[i][j];
  while (K--) {
    int x, y;
    scanf("%d%d", &x, &y);
    top = 0;
    add(x, y);
    add(x, y + m);
    if (find(id[x][y]) == find(id[x][y + m])) {
      for (; top; --top) f[buff[top]] = tmp[top];
      v[x][y] = v[x][y + m] = 0;
    } else
      ++ans;
  }
  printf("%d\n", ans);
  return 0;
}
