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;
  using ll = long long;
  using P = pair<ll, ll>;
  using PP = pair<ll, P>;
  const ll n_ = 2222 + 100, inf = 1e18 + 1, mod = 1e9 + 7, sqrtN = 333;
  ll dy[4] = {-1, 0, 1, 0}, dx[4] = {0, 1, 0, -1};
  ll n, m, k, tc = 1, a, b, c, sum, x, y, z, w, base, ans;
  ll gcd(ll x, ll y) {
    if (!y) return x;
    return gcd(y, x % y);
  }
  ll dp[n_][n_][2], A[n_][n_];
  void solve() {
    ll q;
    cin >> n >> m >> q;
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= m; j++) A[i][j] = 1;
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= m; j++) {
        dp[i][j][0] = A[i][j] * (dp[i][j - 1][1] + A[i][j - 1]);
        dp[i][j][1] = A[i][j] * (dp[i - 1][j][0] + A[i - 1][j]);
        ans += dp[i][j][0] + dp[i][j][1] + A[i][j];
      }
    ll i, j;
    while (q--) {
      cin >> i >> j;
      ans -= dp[i][j][0] + dp[i][j][1] + A[i][j];
      A[i][j] ^= 1;
      dp[i][j][0] = A[i][j] * (dp[i][j - 1][1] + A[i][j - 1]);
      dp[i][j][1] = A[i][j] * (dp[i - 1][j][0] + A[i - 1][j]);
      ans += dp[i][j][0] + dp[i][j][1] + A[i][j];
      b = i + 1, a = j, c = 0;
      while (1) {
        if (b > n || a > m) break;
        ans -= dp[b][a][0] + dp[b][a][1] + A[b][a];
        dp[b][a][0] = A[b][a] * (dp[b][a - 1][1] + A[b][a - 1]);
        dp[b][a][1] = A[b][a] * (dp[b - 1][a][0] + A[b - 1][a]);
        ans += dp[b][a][0] + dp[b][a][1] + A[b][a];
        if (c)
          b++;
        else
          a++;
        c = (c + 1) % 2;
      }
      b = i, a = j + 1, c = 1;
      while (1) {
        if (b > n || a > m) break;
        ans -= dp[b][a][0] + dp[b][a][1] + A[b][a];
        dp[b][a][0] = A[b][a] * (dp[b][a - 1][1] + A[b][a - 1]);
        dp[b][a][1] = A[b][a] * (dp[b - 1][a][0] + A[b - 1][a]);
        ans += dp[b][a][0] + dp[b][a][1] + A[b][a];
        if (c)
          b++;
        else
          a++;
        c = (c + 1) % 2;
      }
      cout << ans << '\n';
    }
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    while (tc--) solve();
  }
  