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 int maxn = 1e3 + 5;
  const long long mod = 998244353;
  const double eps = 1e-10;
  int n, m, q;
  long long dp[maxn][maxn][2];
  int vis[maxn][maxn];
  long long ans;
  long long dfs(int x, int y, int sgn) {
    if (dp[x][y][sgn]) return dp[x][y][sgn];
    dp[x][y][sgn] = 1;
    if (!sgn) {
      if (y < m) dp[x][y][sgn] += dfs(x, y + 1, sgn ^ 1);
    } else {
      if (x < n) dp[x][y][sgn] += dfs(x + 1, y, sgn ^ 1);
    }
    return dp[x][y][sgn];
  }
  void update(int x, int y, int sgn) {
    if (!vis[x][y]) {
      ans -= dp[x][y][0] + dp[x][y][1] - 1;
      dp[x][y][0] = dp[x][y][1] = 1;
      if (y < m) dp[x][y][0] += dp[x][y + 1][1];
      if (x < n) dp[x][y][1] += dp[x + 1][y][0];
      ans += dp[x][y][0] + dp[x][y][1] - 1;
    }
    if (!sgn) {
      if (x > 1) update(x - 1, y, 1);
    } else {
      if (y > 1) update(x, y - 1, 0);
    }
  }
  int main() {
    scanf("%d%d%d", &n, &m, &q);
    dfs(1, 1, 0);
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        dfs(i, j, 0);
        dfs(i, j, 1);
        ans += dp[i][j][0] + dp[i][j][1] - 1;
      }
    }
    while (q--) {
      int x, y;
      scanf("%d%d", &x, &y);
      if (vis[x][y]) {
        dp[x][y][0] = dp[x][y][1] = 1;
        if (y < m) dp[x][y][0] += dp[x][y + 1][1];
        if (x < n) dp[x][y][1] += dp[x + 1][y][0];
        ans += dp[x][y][0] + dp[x][y][1] - 1;
        if (y > 1) update(x, y - 1, 0);
        if (x > 1) update(x - 1, y, 1);
        vis[x][y] = 0;
      } else {
        ans -= dp[x][y][0] + dp[x][y][1] - 1;
        dp[x][y][0] = dp[x][y][1] = 0;
        if (y > 1) update(x, y - 1, 0);
        if (x > 1) update(x - 1, y, 1);
        vis[x][y] = 1;
      }
      printf("%lld\n", ans);
    }
    return 0;
  }
  