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 maxn = 1e6 + 9;
  const long long MOD = 1e9 + 7;
  const int mod = 998244353;
  inline long long qpow(long long b, long long e, long long m) {
    long long a = 1;
    for (; e; e >>= 1, b = b * b % m)
      if (e & 1) a = a * b % m;
    return a;
  }
  long long exgcd(long long a, long long b, long long &x, long long &y) {
    if (b == 0) {
      x = 1, y = 0;
      return a;
    }
    long long d = exgcd(b, a % b, x, y);
    long long z = x;
    x = y, y = z - y * (a / b);
    return d;
  }
  long long dp[1009][1009][3];
  bool a[1009][1009];
  int main() {
    int n, m, q;
    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] = 1, dp[i][j][1] = 1;
      }
    }
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        dp[i][j][0] += dp[i - 1][j][1];
        dp[i][j][1] += dp[i][j - 1][0];
        ans += dp[i][j][0] + dp[i][j][1];
      }
    }
    ans -= n * m;
    while (q--) {
      int x, y;
      scanf("%d%d", &x, &y);
      a[x][y] ^= 1;
      if (a[x][y] == 1) {
        int i = x, j = y, num = 0;
        long long tm = dp[i][j][0];
        while (i <= n && j <= m) {
          if (num % 2 == 0)
            dp[i][j][0] -= tm, j++;
          else
            dp[i][j][1] -= tm, i++;
          ans -= tm;
          num++;
          if (a[i][j] == 1) break;
        }
        i = x, j = y, num = 0;
        tm = dp[i][j][1];
        while (i <= n && j <= m) {
          if (num % 2 == 0)
            dp[i][j][1] -= tm, i++;
          else
            dp[i][j][0] -= tm, j++;
          ans -= tm;
          num++;
          if (a[i][j] == 1) break;
        }
        ans++;
      } else {
        dp[x][y][0] = dp[x - 1][y][1] + 1;
        dp[x][y][1] = dp[x][y - 1][0] + 1;
        int i = x, j = y, num = 1;
        long long tm = dp[x][y][0];
        j++;
        ans += dp[x][y][0];
        while (i <= n && j <= m && a[i][j] == 0) {
          if (num % 2 == 0)
            dp[i][j][0] += tm, j++;
          else
            dp[i][j][1] += tm, i++;
          ans += tm;
          num++;
        }
        i = x, j = y, num = 1;
        tm = dp[x][y][1];
        ans += tm;
        i++;
        while (i <= n && j <= m && a[i][j] == 0) {
          if (num % 2 == 0)
            dp[i][j][1] += tm, i++;
          else
            dp[i][j][0] += tm, j++;
          ans += tm;
          num++;
        }
        ans--;
      }
      printf("%lld\n", ans);
    }
  }
  