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 N = 1005;
  int dx[4] = {1, 0, -1, 0};
  int dy[4] = {0, 1, 0, -1};
  string dirs = "DRUL";
  bool Free[N][N];
  int dir[N][N][4];
  int Freecnt = 0;
  long long sum01 = 0;
  int n, m, q;
  bool isFree(int x, int y) {
    return (1 <= x && x <= n && 1 <= y && y <= m && Free[x][y]);
  }
  int cnt(int x, int y, int dir) {
    if (!isFree(x, y)) return 0;
    return 1 + cnt(x + dx[dir], y + dy[dir], dir ^ 1);
  }
  void add(int x, int y, int dir, int dz) {
    if (!isFree(x, y)) return;
    ::dir[x][y][dir ^ 2] += dz;
    if (dir >= 2) sum01 += dz;
    return add(x + dx[dir], y + dy[dir], dir ^ 1, dz);
  }
  void flip(int x, int y) {
    if (Free[x][y] == 0) {
      Free[x][y] = 1;
      Freecnt++;
      for (int dd = 0; dd < 4; dd++) {
        int cc = cnt(x, y, dd);
        add(x, y, dd ^ 3, cc);
      }
    } else {
      for (int dd = 0; dd < 4; dd++) {
        int cc = cnt(x, y, dd);
        add(x, y, dd ^ 3, -cc);
      }
      Free[x][y] = 0;
      Freecnt--;
    }
  }
  void init() {
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        Free[i][j] = 1;
        Freecnt++;
        dir[i][j][2] = dir[i][j][3] = 1;
        if (i) dir[i][j][2] += dir[i - 1][j][3];
        if (j) dir[i][j][3] += dir[i][j - 1][2];
      }
    }
    for (int i = n; i >= 1; i--) {
      for (int j = m; j >= 1; j--) {
        dir[i][j][0] = dir[i][j][1] = 1;
        if (i < n) dir[i][j][0] += dir[i + 1][j][1];
        if (j < m) dir[i][j][1] += dir[i][j + 1][0];
        sum01 += dir[i][j][0] + dir[i][j][1];
      }
    }
  }
  void all() {
    cout << sum01 << " " << Freecnt << " " << sum01 - Freecnt << endl;
    for (int d = 0; d < 4; d++) {
      cout << d << " -----> " << endl;
      for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
          cout << dir[i][j][d] << " ";
        }
        cout << endl;
      }
    }
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> n >> m >> q;
    init();
    while (q--) {
      int x, y;
      cin >> x >> y;
      flip(x, y);
      cout << sum01 - Freecnt << "\n";
      cout.flush();
    }
  }
  