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>
  #pragma GCC optimize(2, 3, "Ofast")
  using namespace std;
  template <typename T1, typename T2>
  void ckmin(T1 &a, T2 b) {
    if (a > b) a = b;
  }
  template <typename T1, typename T2>
  void ckmax(T1 &a, T2 b) {
    if (a < b) a = b;
  }
  int read() {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= ch == '-', ch = getchar();
    while (isdigit(ch)) x = 10 * x + ch - '0', ch = getchar();
    return f ? -x : x;
  }
  template <typename T>
  void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x / 10);
    putchar(x % 10 + '0');
  }
  template <typename T>
  void print(T x, char let) {
    print(x), putchar(let);
  }
  long long can[1005][1005];
  int a[1005][1005];
  int n, m, q;
  long long ans;
  int main() {
    n = read(), m = read(), q = read();
    for (int i = n; i >= 1; i--) {
      for (int j = m; j >= 1; j--) {
        can[i][j] = can[i + 1][j + 1] + 1 + (i < n) + (j < m) + (i < n && j < m);
        ans += can[i][j];
      }
    }
    while (q--) {
      int x = read(), y = read();
      if (a[x][y]) {
        a[x][y] = 0;
        int x0 = 0, y0 = 0;
        {
          int zx = x, zy = y;
          int dir = 0;
          while (zx >= 1 && zy >= 1 && !a[zx][zy]) {
            x0++;
            if (dir)
              zx--;
            else
              zy--;
            dir ^= 1;
          }
        }
        {
          int zx = x, zy = y;
          int dir = 1;
          while (zx <= n && zy <= m && !a[zx][zy]) {
            y0++;
            if (dir)
              zx++;
            else
              zy++;
            dir ^= 1;
          }
        }
        ans += 1ll * x0 * y0;
        x0 = 0, y0 = 0;
        {
          int zx = x, zy = y;
          int dir = 1;
          while (zx >= 1 && zy >= 1 && !a[zx][zy]) {
            x0++;
            if (dir)
              zx--;
            else
              zy--;
            dir ^= 1;
          }
        }
        {
          int zx = x, zy = y;
          int dir = 0;
          while (zx <= n && zy <= m && !a[zx][zy]) {
            y0++;
            if (dir)
              zx++;
            else
              zy++;
            dir ^= 1;
          }
        }
        ans += 1ll * x0 * y0 - 1;
      } else {
        int x0 = 0, y0 = 0;
        {
          int zx = x, zy = y;
          int dir = 0;
          while (zx >= 1 && zy >= 1 && !a[zx][zy]) {
            x0++;
            if (dir)
              zx--;
            else
              zy--;
            dir ^= 1;
          }
        }
        {
          int zx = x, zy = y;
          int dir = 1;
          while (zx <= n && zy <= m && !a[zx][zy]) {
            y0++;
            if (dir)
              zx++;
            else
              zy++;
            dir ^= 1;
          }
        }
        ans -= 1ll * x0 * y0;
        x0 = 0, y0 = 0;
        {
          int zx = x, zy = y;
          int dir = 1;
          while (zx >= 1 && zy >= 1 && !a[zx][zy]) {
            x0++;
            if (dir)
              zx--;
            else
              zy--;
            dir ^= 1;
          }
        }
        {
          int zx = x, zy = y;
          int dir = 0;
          while (zx <= n && zy <= m && !a[zx][zy]) {
            y0++;
            if (dir)
              zx++;
            else
              zy++;
            dir ^= 1;
          }
        }
        ans -= 1ll * x0 * y0 - 1;
        a[x][y] = 1;
      }
      print(ans, '\n');
    }
  }
  