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 gcd(long long a, long long b) { return a % b ? gcd(b, a % b) : b; }
  const long long mod = 1e9 + 7;
  const long long inf = 0x3f3f3f3f;
  long long qpow(long long a, long long b) {
    long long t = 1;
    while (b != 0) {
      if (b & 1) t = (t * a) % mod;
      a = (a * a) % mod;
      b >>= 1;
    }
    return t;
  }
  template <typename T>
  void re(T& a) {
    for (auto& t : a) cin >> t;
  }
  template <typename T>
  void pr(T& a) {
    for (auto& t : a) cout << t << " ";
    cout << endl;
  }
  int n, m, q;
  const int N = 1005;
  int a[N][N];
  int b[N][N];
  int dfs(int x, int y, int d) {
    if (!a[x][y]) return 0;
    if (d == 0) {
      return 1 + dfs(x - 1, y, 2);
    } else if (d == 1) {
      return 1 + dfs(x + 1, y, 3);
    } else if (d == 2) {
      return 1 + dfs(x, y - 1, 0);
    } else {
      return 1 + dfs(x, y + 1, 1);
    }
  }
  void solve() {
    cin >> n >> m >> q;
    long long ans = 0;
    for (int x = 1; x <= n; ++x) {
      for (int y = 1; y <= m; ++y) {
        a[x][y] = 1;
      }
    }
    long long nm = max(n, m);
    b[1][1] = 1;
    for (int i = 1; i <= nm; ++i) {
      b[i][i] = b[i][i - 1] + 2 * i * i - 1;
      for (int j = i + 1; j <= nm; ++j) {
        b[i][j] = b[j][i] = b[i][j - 1] + 2 * i * i;
      }
    }
    ans = b[n][m];
    while (q--) {
      int x, y;
      cin >> x >> y;
      int f = a[x][y];
      a[x][y] = 1;
      int U = dfs(x, y, 0);
      int D = dfs(x, y, 1);
      int L = dfs(x, y, 2);
      int R = dfs(x, y, 3);
      long long tmp = U * R + D * L - 1;
      ans += f ? -tmp : tmp;
      cout << ans << endl;
      a[x][y] = f ^ 1;
    }
  }
  signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int T = 1;
    while (T--) {
      solve();
    }
    return 0;
  }
  