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;
  inline long long qmi(long long a, long long b, long long mod) {
    long long ans = 1 % mod;
    while (b) {
      if (b & 1) ans = ans * a % mod;
      b >>= 1;
      a = a * a % mod;
    }
    return ans % mod;
  }
  inline long long inv(long long a, long long mod) {
    return qmi(a, mod - 2, mod);
  }
  namespace FastIO {
  char buf[1 << 21], buf2[1 << 21], a[20], *p1 = buf, *p2 = buf, hh = '\n';
  int p, p3 = -1;
  void read() {}
  void print() {}
  inline int getc() {
    return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
               ? EOF
               : *p1++;
  }
  inline void flush() { fwrite(buf2, 1, p3 + 1, stdout), p3 = -1; }
  template <typename T, typename... T2>
  inline void read(T &x, T2 &...oth) {
    int f = 0;
    x = 0;
    char ch = getc();
    while (!isdigit(ch)) {
      if (ch == '-') f = 1;
      ch = getc();
    }
    while (isdigit(ch)) {
      x = x * 10 + ch - 48;
      ch = getc();
    }
    x = f ? -x : x;
    read(oth...);
  }
  template <typename T, typename... T2>
  inline void print(T x, T2... oth) {
    if (p3 > 1 << 20) flush();
    if (x < 0) buf2[++p3] = 45, x = -x;
    do {
      a[++p] = x % 10 + 48;
    } while (x /= 10);
    do {
      buf2[++p3] = a[p];
    } while (--p);
    buf2[++p3] = hh;
    print(oth...);
  }
  }  // namespace FastIO
  const int N = 1010;
  int g[N][N];
  int n, m, q;
  long long ans = 0;
  int dp1[N][N], dp2[N][N];
  void change(int i, int j) {
    while (i <= n && j <= m) {
      if (i < n) {
        ans -= dp1[i + 1][j];
        dp1[i + 1][j] = g[i + 1][j] * (1 + dp2[i][j]);
        ans += dp1[i + 1][j];
      }
      if (j < m) {
        ans -= dp2[i][j + 1];
        dp2[i][j + 1] = g[i][j + 1] * (1 + dp1[i][j]);
        ans += dp2[i][j + 1];
      }
      if (i < n && j < m) {
        ans -= dp1[i + 1][j + 1] + dp2[i + 1][j + 1];
        dp1[i + 1][j + 1] = g[i + 1][j + 1] * (1 + dp2[i][j + 1]);
        dp2[i + 1][j + 1] = g[i + 1][j + 1] * (1 + dp1[i + 1][j]);
        ans += dp1[i + 1][j + 1] + dp2[i + 1][j + 1];
      }
      i++, j++;
    }
  }
  int main() {
    cin >> n >> m >> q;
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= m; j++) {
        dp1[i][j] = dp2[i][j] = 1;
        g[i][j] = 1;
      }
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= m; j++) {
        if (i > 1) dp1[i][j] = dp2[i - 1][j] + 1;
        if (j > 1) dp2[i][j] = dp1[i][j - 1] + 1;
        ans += dp1[i][j] + dp2[i][j];
      }
    int base = n * m;
    while (q--) {
      int x, y;
      cin >> x >> y;
      g[x][y] ^= 1;
      if (!g[x][y]) {
        base--;
        ans -= (dp1[x][y] + dp2[x][y]);
        dp1[x][y] = dp2[x][y] = 0;
      } else {
        base++;
        ans -= (dp1[x][y] + dp2[x][y]);
        if (x > 1)
          dp1[x][y] = dp2[x - 1][y] + 1;
        else
          dp1[x][y] = 1;
        if (y > 1)
          dp2[x][y] = dp1[x][y - 1] + 1;
        else
          dp2[x][y] = 1;
        ans += dp1[x][y] + dp2[x][y];
      }
      change(x, y);
      cout << ans - base << endl;
    }
    return 0;
  }
  