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;
  namespace IO {
  long long len = 0;
  char ibuf[(1 << 20) + 1], *iS, *iT, out[(1 << 25) + 1];
  inline long long read() {
    char ch = (iS == iT ? iT = (iS = ibuf) + fread(ibuf, 1, (1 << 20) + 1, stdin),
               (iS == iT ? EOF : *iS++) : *iS++);
    long long x = 0;
    char t = 0;
    while (ch < '0' || ch > '9')
      t |= ch == '-', ch = (iS == iT ? iT = (iS = ibuf) +
                                            fread(ibuf, 1, (1 << 20) + 1, stdin),
                            (iS == iT ? EOF : *iS++) : *iS++);
    while (ch >= '0' && ch <= '9')
      x = x * 10 + (ch ^ 48),
      ch = (iS == iT ? iT = (iS = ibuf) + fread(ibuf, 1, (1 << 20) + 1, stdin),
            (iS == iT ? EOF : *iS++) : *iS++);
    return t ? -x : x;
  }
  inline void putc(char ch) { out[len++] = ch; }
  template <class T>
  inline void write(T x) {
    if (x < 0) putc('-'), x = -x;
    if (x > 9) write(x / 10);
    out[len++] = x % 10 + 48;
  }
  string getstr(void) {
    string s = "";
    char c = (iS == iT ? iT = (iS = ibuf) + fread(ibuf, 1, (1 << 20) + 1, stdin),
              (iS == iT ? EOF : *iS++) : *iS++);
    while (c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == EOF)
      c = (iS == iT ? iT = (iS = ibuf) + fread(ibuf, 1, (1 << 20) + 1, stdin),
           (iS == iT ? EOF : *iS++) : *iS++);
    while (!(c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == EOF))
      s.push_back(c),
          c = (iS == iT ? iT = (iS = ibuf) + fread(ibuf, 1, (1 << 20) + 1, stdin),
               (iS == iT ? EOF : *iS++) : *iS++);
    return s;
  }
  void putstr(string str, long long begin = 0, long long end = -1) {
    if (end == -1) end = str.size();
    for (long long i = begin; i < end; i++) putc(str[i]);
    return;
  }
  inline void flush() {
    fwrite(out, 1, len, stdout);
    len = 0;
  }
  }  // namespace IO
  using IO::flush;
  using IO::getstr;
  using IO::putc;
  using IO::putstr;
  using IO::read;
  using IO::write;
  const long long N = 3000;
  long long n, m, q, ans, cnt, a[N][N], dp1[N][N], dp2[N][N];
  void solve() {
    n = read(), m = read(), q = read();
    for (long long i = 1; i <= n; i++)
      for (long long j = 1; j <= m; j++) dp1[i][j] = dp2[i][j] = 1;
    for (long long i = 1; i <= n; i++) {
      for (long long j = 1; j <= m; j++) {
        if (i > 1) dp1[i][j] = 1 + dp2[i - 1][j];
        if (j > 1) dp2[i][j] = 1 + dp1[i][j - 1];
        a[i][j] = 1;
      }
    }
    long long ans = 0, tot = n * m;
    for (long long i = 1; i <= n; i++)
      for (long long j = 1; j <= m; j++) ans += dp1[i][j] + dp2[i][j];
    while (q--) {
      long long i, j;
      i = read(), j = read();
      a[i][j] ^= 1;
      if (a[i][j] == 0) {
        tot--;
        ans -= (dp1[i][j] + dp2[i][j]);
        dp1[i][j] = dp2[i][j] = 0;
        ans += (dp1[i][j] + dp2[i][j]);
      } else {
        tot++;
        ans -= dp1[i][j] + dp2[i][j];
        if (i > 1)
          dp1[i][j] = 1 + dp2[i - 1][j];
        else
          dp1[i][j] = 1;
        if (j > 1)
          dp2[i][j] = 1 + dp1[i][j - 1];
        else
          dp2[i][j] = 1;
        ans += dp1[i][j] + dp2[i][j];
      }
      while (i <= n && j <= m) {
        if (i < n) {
          ans -= dp1[i + 1][j];
          dp1[i + 1][j] = a[i + 1][j] * (1 + dp2[i][j]);
          ans += dp1[i + 1][j];
        }
        if (j < m) {
          ans -= dp2[i][j + 1];
          dp2[i][j + 1] = a[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] = a[i + 1][j + 1] * (1 + dp2[i][j + 1]);
          dp2[i + 1][j + 1] = a[i + 1][j + 1] * (1 + dp1[i + 1][j]);
          ans += dp1[i + 1][j + 1] + dp2[i + 1][j + 1];
        }
        i++, j++;
      }
      cout << ans - tot << " " << endl;
    }
  }
  signed main() {
    solve();
    return 0;
  }
  