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;
  int q, n, m;
  struct node {
    int x, y;
  };
  queue<node> Q;
  int a[1005][1005];
  int vis[1005][1005];
  int use, nuse;
  int dp[1005][1005][2];
  int bfs(int x, int y, int op, int z) {
    int qwq = 0;
    Q.push({x, y});
    while (Q.size()) {
      node v = Q.front();
      Q.pop();
      if (op) {
        v.x += z;
        op ^= 1;
      } else {
        v.y += z;
        op ^= 1;
      }
      if (v.x >= 1 && v.x <= n && v.y >= 1 && v.y <= m) {
        if (!vis[v.x][v.y]) {
          qwq++;
          Q.push({v.x, v.y});
        } else
          return qwq;
      }
    }
    return qwq;
  }
  int main() {
    long long ans = 0;
    scanf("%d%d%d", &n, &m, &q);
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= m; ++j) {
        dp[i][j][0] = dp[i][j][1] = 1;
      }
    }
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= m; ++j) {
        dp[i][j][0] += dp[i][j - 1][1];
        dp[i][j][1] += dp[i - 1][j][0];
      }
    }
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= m; ++j) {
        dp[i][j][0]--;
        dp[i][j][1]--;
        ans += dp[i][j][0] + dp[i][j][1] + 1;
      }
    }
    int x, y;
    long long now = ans;
    while (q--) {
      scanf("%d%d", &x, &y);
      int a1 = bfs(x, y, 0, 1);
      int a2 = bfs(x, y, 1, 1);
      int b1 = bfs(x, y, 0, -1);
      int b2 = bfs(x, y, 1, -1);
      if (!vis[x][y]) {
        now -= ((a1 * b2) + (a2 * b1) + a1 + b1 + a2 + b2);
        now--;
        vis[x][y] = 1;
        printf("%lld\n", now);
      } else {
        now += (a1 * b2) + (a2 * b1) + a1 + b1 + a2 + b2;
        now++;
        vis[x][y] = 0;
        printf("%lld\n", now);
      }
    }
    return 0;
  }
  