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 N = 1000 + 10, M = 998244353;
  int t;
  int n, m, q;
  bool locked[N][N];
  long long num[N][N][2];
  long long ans = 0;
  int calc(int i, int j, int t) {
    if (i > n || j > m) return 0;
    if (num[i][j][t] != 0) return num[i][j][t];
    num[i][j][t] = 1;
    if (t == 0)
      num[i][j][t] += calc(i + 1, j, t ^ 1);
    else
      num[i][j][t] += calc(i, j + 1, t ^ 1);
    return num[i][j][t];
  }
  void go(int i, int j, int t, int val) {
    if (i <= 0 || j <= 0 || locked[i][j]) return;
    if (t == 0) {
      num[i][j][t] -= val;
      ans -= val;
      return go(i, j - 1, t ^ 1, val);
    } else {
      num[i][j][t] -= val;
      ans -= val;
      return go(i - 1, j, t ^ 1, val);
    }
  }
  void go2(int i, int j, int t) {
    num[i][j][t] = 1;
    ans++;
    if (t == 0) {
      if (i < n) {
        num[i][j][t] += num[i + 1][j][t ^ 1];
        ans += num[i + 1][j][t ^ 1];
      }
    } else {
      if (j < m) {
        num[i][j][t] += num[i][j + 1][t ^ 1];
        ans += num[i][j + 1][t ^ 1];
      }
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> n >> m >> q;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        for (int k = 0; k < 2; k++) ans += calc(i, j, k);
      }
    }
    long long free = n * m;
    while (q--) {
      int x, y;
      cin >> x >> y;
      if (!locked[x][y]) {
        go(x, y, 0, num[x][y][0]);
        go(x, y, 1, num[x][y][1]);
        locked[x][y] = 1;
        free--;
      } else {
        go2(x, y, 0);
        go2(x, y, 1);
        locked[x][y] = 0;
        go(x, y, 0, -num[x][y][0]);
        go(x, y, 1, -num[x][y][1]);
        num[x][y][0] /= 2;
        num[x][y][1] /= 2;
        ans -= num[x][y][0] + num[x][y][1];
        free++;
      }
      cout << ans - free << "\n";
    }
    return 0;
  }
  