description: You are given a matrix, consisting of n rows and m columns. The rows are numbered top to bottom, the columns are numbered left to right.
  
  Each cell of the matrix can be either free or locked.
  
  Let's call a path in the matrix a staircase if it: 
  
    * starts and ends in the free cell; 
    * visits only free cells; 
    * has one of the two following structures: 
      1. the second cell is 1 to the right from the first one, the third cell is 1 to the bottom from the second one, the fourth cell is 1 to the right from the third one, and so on; 
      2. the second cell is 1 to the bottom from the first one, the third cell is 1 to the right from the second one, the fourth cell is 1 to the bottom from the third one, and so on. 
  
  
  
  In particular, a path, consisting of a single cell, is considered to be a staircase.
  
  Here are some examples of staircases:
  
  <image>
  
  Initially all the cells of the matrix are free.
  
  You have to process q queries, each of them flips the state of a single cell. So, if a cell is currently free, it makes it locked, and if a cell is currently locked, it makes it free.
  
  Print the number of different staircases after each query. Two staircases are considered different if there exists such a cell that appears in one path and doesn't appear in the other path.
  
  Input
  
  The first line contains three integers n, m and q (1 ≤ n, m ≤ 1000; 1 ≤ q ≤ 10^4) — the sizes of the matrix and the number of queries.
  
  Each of the next q lines contains two integers x and y (1 ≤ x ≤ n; 1 ≤ y ≤ m) — the description of each query.
  
  Output
  
  Print q integers — the i-th value should be equal to the number of different staircases after i queries. Two staircases are considered different if there exists such a cell that appears in one path and doesn't appear in the other path.
  
  Examples
  
  Input
  
  
  2 2 8
  1 1
  1 1
  1 1
  2 2
  1 1
  1 2
  2 1
  1 1
  
  
  Output
  
  
  5
  10
  5
  2
  5
  3
  1
  0
  
  
  Input
  
  
  3 4 10
  1 4
  1 2
  2 3
  1 2
  2 3
  3 2
  1 3
  3 4
  1 3
  3 1
  
  
  Output
  
  
  49
  35
  24
  29
  49
  39
  31
  23
  29
  27
  
  
  Input
  
  
  1000 1000 2
  239 634
  239 634
  
  
  Output
  
  
  1332632508
  1333333000
solution: #include <bits/stdc++.h>
  using namespace std;
  const long long mod = 1000000007;
  const long long infinity = 1000000000000000000;
  const int inf = 1e9 + 5;
  const int maxn = 1005;
  int n, m, q;
  int state[maxn][maxn] = {0};
  long long dp[maxn][maxn][2] = {0};
  long long ans = 0;
  int in(int x, int y) {
    if (x >= 0 && x < n && y >= 0 && y < m) {
      return 1;
    }
    return 0;
  }
  void g(int x, int y) {
    if (in(x + 1, y)) {
      ans -= dp[x][y][1];
      int extra = state[x + 1][y];
      dp[x][y][1] = extra + dp[x + 1][y][0];
      ans += dp[x][y][1];
    }
    if (in(x, y + 1)) {
      ans -= dp[x][y][0];
      int extra = state[x][y + 1];
      dp[x][y][0] = extra + dp[x][y + 1][1];
      ans += dp[x][y][0];
    }
  }
  void f(int x, int y) {
    if (!in(x, y)) {
      return;
    }
    if (!state[x][y]) {
      return;
    }
    g(x, y);
    if (in(x - 1, y) && state[x - 1][y]) {
      g(x - 1, y);
    }
    if (in(x, y - 1) && state[x][y - 1]) {
      g(x, y - 1);
    }
    f(x - 1, y - 1);
  }
  void solve() {
    cin >> n >> m >> q;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        state[i][j] = 1;
      }
    }
    for (int i = n - 1; i >= 0; i--) {
      for (int j = m - 1; j >= 0; j--) {
        g(i, j);
        ans++;
      }
    }
    for (int Q = 0; Q < q; Q++) {
      int x, y;
      cin >> x >> y;
      x--;
      y--;
      state[x][y] ^= 1;
      if (!state[x][y]) {
        ans -= dp[x][y][0];
        ans -= dp[x][y][1];
        dp[x][y][0] = 0;
        dp[x][y][1] = 0;
        if (in(x - 1, y) && state[x - 1][y]) {
          g(x - 1, y);
        }
        if (in(x, y - 1) && state[x][y - 1]) {
          g(x, y - 1);
        }
        ans--;
        f(x - 1, y - 1);
      } else {
        ans++;
        f(x, y);
      }
      cout << ans << "\n";
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int TC = 1;
    for (int tc = 1; tc <= TC; tc++) {
      solve();
    }
    return 0;
  }
  