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;
  using ll = long long;
  using ull = unsigned long long;
  using ld = long double;
  using db = double;
  using str = string;
  using pi = pair<int, int>;
  using pl = pair<ll, ll>;
  using pd = pair<db, db>;
  using vi = vector<int>;
  using vb = vector<bool>;
  using vl = vector<ll>;
  using vd = vector<db>;
  using vs = vector<str>;
  using vpi = vector<pi>;
  using vpl = vector<pl>;
  using vpd = vector<pd>;
  template <class T>
  using V = vector<T>;
  template <class T, size_t SZ>
  using AR = array<T, SZ>;
  template <class T>
  using PR = pair<T, T>;
  template <class T>
  int lwb(V<T>& a, const T& b) {
    return int(lower_bound(begin(a), end(a), b) - begin(a));
  }
  int g[1001][1001], c[1001][1001][2][2], d[1001][1001][3];
  ;
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int n, m, qc;
    cin >> n >> m >> qc;
    memset(g, 0, sizeof(g));
    memset(c, 0, sizeof(c));
    memset(d, 0, sizeof(d));
    vector<ll> ans;
    vector<array<int, 2>> q(qc);
    ll total = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        d[i][j][0] = 1;
        if (i == 1) {
          if (j > 1) {
            d[i][j][1] = 1;
          }
        } else {
          d[i][j][2] = d[i - 1][j][0] + d[i - 1][j][1];
          if (j > 1) {
            d[i][j][1] = d[i][j - 1][0] + d[i][j - 1][2];
          }
        }
        total += d[i][j][0] + d[i][j][1] + d[i][j][2];
      }
    }
    for (int i = (0); i < (qc); ++i) {
      int r, c;
      cin >> r >> c;
      if (g[r][c]) {
        g[r][c] = 0;
        int x = r, y = c;
        ll cntu = 0, cntd = 0;
        x = r, y = c;
        while (true) {
          x--;
          if (x < 1) break;
          if (g[x][y]) break;
          cntu++;
          y--;
          if (y < 1) break;
          if (g[x][y]) break;
          cntu++;
        }
        x = r, y = c;
        while (true) {
          y++;
          if (y > m) break;
          if (g[x][y]) break;
          cntd++;
          x++;
          if (x > n) break;
          if (g[x][y]) break;
          cntd++;
        }
        total += (1 + cntu) * (1 + cntd);
        cntu = cntd = 0;
        x = r, y = c;
        while (true) {
          y--;
          if (y < 1) break;
          if (g[x][y]) break;
          cntu++;
          x--;
          if (x < 1) break;
          if (g[x][y]) break;
          cntu++;
        }
        x = r, y = c;
        while (true) {
          x++;
          if (x > n) break;
          if (g[x][y]) break;
          cntd++;
          y++;
          if (y > m) break;
          if (g[x][y]) break;
          cntd++;
        }
        total += (1 + cntu) * (1 + cntd) - 1;
        ans.push_back(total);
      } else {
        g[r][c] = 1;
        int x = r, y = c;
        ll cntu = 0, cntd = 0;
        x = r, y = c;
        while (true) {
          x--;
          if (x < 1) break;
          if (g[x][y]) break;
          cntu++;
          y--;
          if (y < 1) break;
          if (g[x][y]) break;
          cntu++;
        }
        x = r, y = c;
        while (true) {
          y++;
          if (y > m) break;
          if (g[x][y]) break;
          cntd++;
          x++;
          if (x > n) break;
          if (g[x][y]) break;
          cntd++;
        }
        total -= (1 + cntu) * (1 + cntd);
        cntu = cntd = 0;
        x = r, y = c;
        while (true) {
          y--;
          if (y < 1) break;
          if (g[x][y]) break;
          cntu++;
          x--;
          if (x < 1) break;
          if (g[x][y]) break;
          cntu++;
        }
        x = r, y = c;
        while (true) {
          x++;
          if (x > n) break;
          if (g[x][y]) break;
          cntd++;
          y++;
          if (y > m) break;
          if (g[x][y]) break;
          cntd++;
        }
        total -= (1 + cntu) * (1 + cntd) - 1;
        ans.push_back(total);
      }
    }
    for (int i = (0); i < (qc); ++i) {
      cout << ans[i] << endl;
    }
  }
  