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;
  long long int arr[1001][1001];
  long long int dp[1001][1001];
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long int n, m, q;
    cin >> n >> m >> q;
    for (int i = 0; i <= n; i++)
      for (int j = 0; j <= m; j++) arr[i][j] = 1, dp[i][j] = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        long long int d = n - i, r = m - j;
        arr[i][j] += (4 * min(d, r));
        if (d != r) arr[i][j]++;
      }
    }
    long long int sum = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        sum += arr[i][j];
      }
    }
    while (q--) {
      long long int x, y;
      cin >> x >> y;
      long long int r = x, c = y;
      if (dp[r][c] == 0) {
        long long int temp = 0;
        long long int cnt = 0;
        while (r <= n && c <= m && dp[r][c] == 0) {
          temp++;
          if (cnt % 2 == 0)
            r++;
          else
            c++;
          cnt++;
        }
        r = x, c = y;
        cnt = 0;
        while (r >= 1 && c >= 1 && dp[r][c] == 0) {
          arr[r][c] -= temp;
          sum -= temp;
          if (cnt % 2 == 0)
            c--;
          else
            r--;
          cnt++;
        }
        temp = 0;
        cnt = 0;
        r = x, c = y;
        while (r <= n && c <= m && dp[r][c] == 0) {
          temp++;
          if (cnt % 2 == 0)
            c++;
          else
            r++;
          cnt++;
        }
        r = x, c = y;
        cnt = 0;
        while (r >= 1 && c >= 1 && dp[r][c] == 0) {
          arr[r][c] -= temp;
          sum -= temp;
          if (cnt % 2 == 0)
            r--;
          else
            c--;
          cnt++;
        }
        arr[x][y]++;
        sum += 1;
        dp[x][y] = 1;
        cout << sum << endl;
      } else {
        dp[x][y] = 0;
        long long int temp = 0;
        long long int cnt = 0;
        while (r <= n && c <= m && dp[r][c] == 0) {
          temp++;
          if (cnt % 2 == 0)
            r++;
          else
            c++;
          cnt++;
        }
        r = x, c = y;
        cnt = 0;
        while (r >= 1 && c >= 1 && dp[r][c] == 0) {
          arr[r][c] += temp;
          sum += temp;
          if (cnt % 2 == 0)
            c--;
          else
            r--;
          cnt++;
        }
        temp = 0;
        cnt = 0;
        r = x, c = y;
        while (r <= n && c <= m && dp[r][c] == 0) {
          temp++;
          if (cnt % 2 == 0)
            c++;
          else
            r++;
          cnt++;
        }
        r = x, c = y;
        cnt = 0;
        while (r >= 1 && c >= 1 && dp[r][c] == 0) {
          arr[r][c] += temp;
          sum += temp;
          if (cnt % 2 == 0)
            r--;
          else
            c--;
          cnt++;
        }
        arr[x][y]--;
        sum -= 1;
        cout << sum << endl;
      }
    }
  }
  