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;
int R, C, Q, N;
bool isLand[3005][3005];
int dx[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int dy[] = {-1, 0, 1, -1, 1, -1, 0, 1};
int connectedRegion = 0, parent[2][10], par[3000 * 3000 * 2 + 5];
int findpar(int x) { return x == par[x] ? par[x] : par[x] = findpar(par[x]); }
int parIdx(int r, int c) { return (r - 1) * C + c; }
void solve(int r, int c) {
  if (r < 1 || r > R) return;
  bool rot = false;
  if (c < 1) rot = 1, c = C;
  if (c > C) rot = 1, c = 1;
  if (!isLand[r][c]) return;
  connectedRegion++;
  int u = findpar(parIdx(r, c) + rot * N);
  int v = findpar(parIdx(r, c) + (!rot) * N);
  parent[0][connectedRegion] = u;
  parent[1][connectedRegion] = v;
}
bool ok(int r, int c) {
  connectedRegion = 0;
  for (int d = 0; d < 8; d++) solve(r + dx[d], c + dy[d]);
  for (int i = 1; i <= connectedRegion; i++) {
    for (int j = 1; j <= connectedRegion; j++) {
      if (parent[0][i] == parent[1][j]) return false;
    }
  }
  int pu = findpar(parIdx(r, c));
  int pv = findpar(parIdx(r, c) + N);
  for (int i = 1; i <= connectedRegion; i++) {
    par[parent[0][i]] = pu;
    par[parent[1][i]] = pv;
  }
  return isLand[r][c] = 1;
}
int main() {
  ios ::sync_with_stdio(false);
  cin >> R >> C >> Q;
  if (C == 1) {
    cout << 0 << endl;
    return 0;
  }
  N = R * C;
  for (int i = 1; i <= 2 * N; i++) par[i] = i;
  int ans = 0;
  for (int q = 1; q <= Q; q++) {
    int r, c;
    cin >> r >> c;
    ans += ok(r, c);
  }
  cout << ans << endl;
  return 0;
}
