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 = 1005;
  long long aa[N][N];
  long long n, m;
  long long compute() {
    long long ans = 0;
    for (long long i = 1; i <= n; i++) {
      for (long long j = 1; j <= m; j++) {
        long long r = m - j;
        long long d = n - i;
        if (r == d) {
          ans += 4 * r;
        } else if (r > d) {
          ans += 2 * d + 1;
          ans += 2 * d;
        } else {
          ans += 2 * r + 1;
          ans += 2 * r;
        }
      }
    }
    ans += n * m;
    return ans;
  }
  void solve(long long nn) {
    long long q;
    cin >> n >> m >> q;
    long long ans = compute();
    map<long long, set<long long>> mp;
    for (long long i = 0; i < n + 2; i++) {
      for (long long j = 0; j < m + 2; j++) aa[i][j] = 1;
    }
    for (long long i = 1; i < n + 1; i++) {
      for (long long j = 1; j < m + 1; j++) aa[i][j] = 0;
    }
    for (long long i = 0; i < n + 2; i++) {
      for (long long j = 0; j < m + 2; j++) {
        if (aa[i][j]) mp[i - j].insert(i);
      }
    }
    while (q--) {
      long long x, y;
      cin >> x >> y;
      if (aa[x][y]) {
        aa[x][y] = 0;
        mp[x - y].erase(x);
        auto it2 = mp[x - y].lower_bound(x);
        long long piche2 = *it2;
        it2--;
        long long aage2 = *it2;
        auto it3 = mp[x + 1 - y].lower_bound(x + 1);
        long long piche3 = *it3;
        it3--;
        long long aage3 = *it3;
        auto it1 = mp[x - 1 - y].lower_bound(x);
        long long piche1 = *it1;
        it1--;
        long long aage1 = *it1;
        long long dup = x - (aage2)-1;
        long long d3up = x - aage3;
        long long d1up = x - 1 - aage1;
        long long dd = piche2 - x - 1;
        long long d3d = piche3 - x - 1;
        long long d1d = piche1 - x;
        long long a, b;
        if (dup == d3up) {
          a = 2 * dup;
        } else {
          if (dup < d3up)
            a = 2 * min(dup, d3up) + 1;
          else
            a = 2 * min(dup, d3up);
        }
        if (dd == d3d) {
          b = 2 * dd;
        } else {
          if (dd < d3d)
            b = 2 * min(dd, d3d) + 1;
          else
            b = 2 * min(dd, d3d);
        }
        ans += a + b + a * b;
        if (dup == d1up) {
          a = 2 * dup;
        } else {
          if (dup < d1up)
            a = 2 * min(dup, d1up) + 1;
          else
            a = 2 * min(dup, d1up);
        }
        if (dd == d1d) {
          b = 2 * dd;
        } else {
          if (dd < d1d)
            b = 2 * min(dd, d1d) + 1;
          else
            b = 2 * min(dd, d1d);
        }
        ans += a + b + a * b;
        ans++;
      } else {
        auto it2 = mp[x - y].lower_bound(x);
        long long piche2 = *it2;
        it2--;
        long long aage2 = *it2;
        auto it3 = mp[x + 1 - y].lower_bound(x + 1);
        long long piche3 = *it3;
        it3--;
        long long aage3 = *it3;
        auto it1 = mp[x - 1 - y].lower_bound(x);
        long long piche1 = *it1;
        it1--;
        long long aage1 = *it1;
        long long dup = x - (aage2)-1;
        long long d3up = x - aage3;
        long long d1up = x - 1 - aage1;
        long long dd = piche2 - x - 1;
        long long d3d = piche3 - x - 1;
        long long d1d = piche1 - x;
        long long a, b;
        if (dup == d3up) {
          a = 2 * dup;
        } else {
          if (dup < d3up)
            a = 2 * min(dup, d3up) + 1;
          else
            a = 2 * min(dup, d3up);
        }
        if (dd == d3d) {
          b = 2 * dd;
        } else {
          if (dd < d3d)
            b = 2 * min(dd, d3d) + 1;
          else
            b = 2 * min(dd, d3d);
        }
        ans -= (a + b + a * b);
        if (dup == d1up) {
          a = 2 * dup;
        } else {
          if (dup < d1up)
            a = 2 * min(dup, d1up) + 1;
          else
            a = 2 * min(dup, d1up);
        }
        if (dd == d1d) {
          b = 2 * dd;
        } else {
          if (dd < d1d)
            b = 2 * min(dd, d1d) + 1;
          else
            b = 2 * min(dd, d1d);
        }
        ans -= (a + b + a * b);
        ans--;
        aa[x][y] = 1;
        mp[x - y].insert(x);
      }
      cout << ans << "\n";
    }
    return;
  }
  int main() {
    std::cout << std::fixed;
    std::cout << std::setprecision(15);
    ;
    long long t;
    t = 1;
    for (long long i = 1; i <= t; i++) solve(i);
    return 0;
  }
  