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>
  #pragma GCC optimize("Ofast")
  #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
  #pragma GCC optimize("unroll-loops")
  using namespace std;
  template <class T, class second>
  ostream &operator<<(ostream &os, const pair<T, second> &v) {
    os << v.first << ' ' << v.second;
    return os;
  }
  template <class T>
  ostream &operator<<(ostream &os, const vector<T> &v) {
    for (T i : v) os << i << ' ';
    return os;
  }
  template <class T>
  ostream &operator<<(ostream &os, const set<T> &v) {
    for (T i : v) os << i << ' ';
    return os;
  }
  template <class T, class second>
  ostream &operator<<(ostream &os, const map<T, second> &v) {
    for (auto i : v) os << '(' << i.first << "," << i.second << ')' << ' ';
    return os;
  }
  template <class T, class second>
  ostream &operator<<(ostream &os, const unordered_map<T, second> &v) {
    for (auto i : v) os << '(' << i.first << "," << i.second << ')' << ' ';
    return os;
  }
  long long m1(vector<vector<int> > &r, vector<vector<int> > &d,
               vector<vector<int> > &s, int x, int y, long long c, int f,
               int fl) {
    long long t = c;
    bool f2 = true;
    while (x < r.size() && y < r[0].size() && f2) {
      if (f == 1) {
        f--;
        x++;
        if (x < r.size() && s[x][y] == 0) {
          d[x][y] += fl * c;
          t += c;
        }
        if (x < r.size() && s[x][y] == 1) {
          f2 = false;
        }
      } else {
        f++;
        y++;
        if (y < r[0].size() && s[x][y] == 0) {
          t += c;
          r[x][y] += fl * c;
        }
        if (y < r[0].size() && s[x][y] == 1) {
          f2 = false;
        }
      }
    }
    return t;
  }
  void solve() {
    int n, m, q;
    cin >> n >> m >> q;
    long long a = -1 * n * m;
    vector<vector<int> > s(n, vector<int>(m, 0));
    vector<vector<int> > r(n, vector<int>(m, 1)), d(n, vector<int>(m, 1));
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        if (j > 0) {
          r[i][j] += d[i][j - 1];
        }
        if (i > 0) {
          d[i][j] += r[i - 1][j];
        }
        a += r[i][j] + d[i][j];
      }
    }
    while (q--) {
      int x, y;
      cin >> x >> y;
      x--;
      y--;
      if (s[x][y] == 0) {
        s[x][y] = 1;
        long long t = m1(r, d, s, x, y, r[x][y], 1, -1);
        t += m1(r, d, s, x, y, d[x][y], 0, -1);
        t--;
        a -= t;
        r[x][y] = 0;
        d[x][y] = 0;
      } else {
        s[x][y] = 0;
        r[x][y] = 1;
        d[x][y] = 1;
        if (x > 0) {
          d[x][y] += r[x - 1][y];
        }
        if (y > 0) {
          r[x][y] += d[x][y - 1];
        }
        long long t = m1(r, d, s, x, y, r[x][y], 1, 1);
        t += m1(r, d, s, x, y, d[x][y], 0, 1);
        t--;
        a += t;
      }
      cout << a;
      cout << "\n";
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long tt = 1;
    while (tt--) {
      solve();
    }
    return 0;
  }
  