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 long long N = 1e3 + 10;
  long long n, m, Q, res, cnt;
  long long a[N][N], f[N][N];
  inline void query(long long x, long long y, long long tag) {
    if (x ^ n) {
      long long cur = min(f[x + 1][y], f[x][y]);
      res += tag * cur;
      if (f[x + 1][y] >= f[x][y])
        res += max(0LL, f[x][y] - 1) * tag;
      else
        res += tag * f[x + 1][y];
    }
    if (y ^ m) {
      long long cur = min(f[x][y + 1], f[x][y]);
      res += tag * cur;
      if (f[x][y + 1] >= f[x][y])
        res += max(0LL, f[x][y] - 1) * tag;
      else
        res += tag * f[x][y + 1];
    }
    return;
  }
  inline void modify(long long x, long long y) {
    if (a[x][y])
      --cnt;
    else
      ++cnt;
    a[x][y] ^= 1;
    if (a[x][y])
      f[x][y] = f[x + 1][y + 1] + 1;
    else
      f[x][y] = 0;
    return;
  }
  inline void upd(long long x, long long y, long long tag) {
    long long l = x, r = y;
    while (l && r) query(l, r, tag), --l, --r;
    return;
  }
  inline void update(long long x, long long y) {
    upd(x, y, -1), upd(x - 1, y, -1);
    upd(x, y - 1, -1);
    modify(x, y);
    long long l = x - 1, r = y - 1;
    while (l && r) {
      if (a[l][r])
        f[l][r] = f[l + 1][r + 1] + 1;
      else
        f[l][r] = 0;
      --l, --r;
    }
    upd(x, y, 1);
    upd(x - 1, y, 1);
    upd(x, y - 1, 1);
    return;
  }
  signed main() {
    scanf("%lld%lld%lld", &n, &m, &Q);
    cnt = n * m;
    for (long long i = 1; i <= n; ++i)
      for (long long j = 1; j <= m; ++j) a[i][j] = 1;
    for (long long i = n; i; --i)
      for (long long j = m; j; --j) f[i][j] = f[i + 1][j + 1] + 1;
    for (long long i = 1; i <= n; ++i)
      for (long long j = 1; j <= m; ++j) query(i, j, 1);
    while (Q--) {
      long long x, y;
      scanf("%lld%lld", &x, &y);
      update(x, y);
      printf("%lld\n", res + cnt);
    }
    return 0;
  }
  