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;
  void judge() {}
  void fastio() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
  }
  long long int p1 = 1e9 + 7;
  long long int p2 = 998244353;
  struct segtree {
    vector<int> v;
    int n, n2;
    void init(int x) {
      n = x;
      int a = 1;
      while (a <= x) a *= 2;
      a *= 2;
      n2 = a;
      v.assign(n2, 0);
    }
    int return_max(int curr, int r2, int l, int r) {
      if (l == r) {
        return v[curr];
      }
      int mid = (r + l) / 2;
      if (r2 <= mid) return return_max(2 * curr + 1, r2, l, mid);
      return max(v[2 * curr + 1], return_max(2 * curr + 2, r2, mid + 1, r));
    }
    int return_max(int r) {
      if (r < 0) return 0;
      return return_max(0, r, 0, n - 1);
    }
    void update(int curr, int pos, int val, int l, int r) {
      if (l == r) {
        v[curr] = max(v[curr], val);
        return;
      }
      int mid = (l + r) / 2;
      if (pos <= mid)
        update(2 * curr + 1, pos, val, l, mid);
      else
        update(2 * curr + 2, pos, val, mid + 1, r);
      v[curr] = max(v[2 * curr + 1], v[2 * curr + 2]);
    }
    void update(int pos, int val) { update(0, pos, val, 0, n - 1); }
    int ans() {
      int a = 0;
      for (auto val : v) a = max(a, val);
      return a;
    }
  };
  bool compare(pair<long long int, long long int>& a,
               pair<long long int, long long int>& b) {
    if (a.first != b.first) return a.first > b.first;
    return a.second < b.second;
  }
  int find_set(vector<int>& parent, int x) {
    if (parent[x] == x) return x;
    return parent[x] = find_set(parent, parent[x]);
  }
  void union_set(vector<int>& parent, int x, int y) {
    x = find_set(parent, x);
    y = find_set(parent, y);
    if (x != y) {
      parent[x] = y;
    }
  }
  void pre() {}
  void solve2() {
    long long int n, m, q;
    cin >> n >> m >> q;
    vector<vector<pair<pair<long long int, long long int>, long long int> > > v(
        n, vector<pair<pair<long long int, long long int>, long long int> >(m));
    v[0][0] = {{0, 0}, 1};
    long long int ans = 1;
    for (int i = 1; i < m; i++) {
      v[0][i] = {{0, 1}, 1};
      ans += 2;
    }
    for (int i = 1; i < n; i++) {
      v[i][0] = {{1, 0}, 1};
      ans += 2;
    }
    for (int i = 1; i < n; i++) {
      for (int j = 1; j < m; j++) {
        v[i][j] = {{v[i - 1][j].first.second + v[i - 1][j].second,
                    v[i][j - 1].first.first + v[i][j - 1].second},
                   1};
        ans += v[i][j].first.first;
        ans += v[i][j].first.second;
        ans += v[i][j].second;
      }
    }
    for (int i = 0; i < q; i++) {
      long long int x, y;
      cin >> x >> y;
      x--;
      y--;
      if (v[x][y].second == 1) {
        ans -= v[x][y].second;
        ans -= v[x][y].first.second;
        ans -= v[x][y].first.first;
        v[x][y] = {{0, 0}, 0};
      } else {
        if (x == 0) {
          if (y == 0) {
            v[x][y] = {{0, 0}, 1};
          } else {
            v[x][y] = {{0, v[x][y - 1].second}, 1};
          }
        } else {
          if (y == 0) {
            v[x][y] = {{v[x - 1][y].second, 0}, 1};
          } else {
            v[x][y] = {{v[x - 1][y].first.second + v[x - 1][y].second,
                        v[x][y - 1].first.first + v[x][y - 1].second},
                       1};
          }
        }
        ans += v[x][y].first.first;
        ans += v[x][y].first.second;
        ans += v[x][y].second;
      }
      while (x < n && y < m) {
        if (x < (n - 1)) {
          if (v[x + 1][y].second != 0) {
            ans -= v[x + 1][y].first.first;
            ans -= v[x + 1][y].first.second;
            ans -= v[x + 1][y].second;
            if (y == 0) {
              v[x + 1][y] = {{v[x][y].first.second + v[x][y].second, 0}, 1};
            } else {
              v[x + 1][y] = {
                  {v[x][y].first.second + v[x][y].second,
                   v[x + 1][y - 1].first.first + v[x + 1][y - 1].second},
                  1};
            }
            ans += v[x + 1][y].first.first;
            ans += v[x + 1][y].first.second;
            ans += v[x + 1][y].second;
          }
        }
        if (y < (m - 1) && v[x][y + 1].second != 0) {
          ans -= v[x][y + 1].first.first;
          ans -= v[x][y + 1].first.second;
          ans -= v[x][y + 1].second;
          if (x == 0) {
            v[x][y + 1] = {{0, v[x][y].first.first + v[x][y].second}, 1};
          } else {
            v[x][y + 1] = {{v[x - 1][y + 1].first.second + v[x - 1][y + 1].second,
                            v[x][y].first.first + v[x][y].second},
                           1};
          }
          ans += v[x][y + 1].first.first;
          ans += v[x][y + 1].first.second;
          ans += v[x][y + 1].second;
        }
        if (x < (n - 1) && y < (m - 1) && v[x + 1][y + 1].second != 0) {
          ans -= v[x + 1][y + 1].first.first;
          ans -= v[x + 1][y + 1].first.second;
          ans -= v[x + 1][y + 1].second;
          v[x + 1][y + 1] = {{v[x][y + 1].first.second + v[x][y + 1].second,
                              v[x + 1][y].first.first + v[x + 1][y].second},
                             1};
          ans += v[x + 1][y + 1].first.first;
          ans += v[x + 1][y + 1].first.second;
          ans += v[x + 1][y + 1].second;
        }
        x += 1;
        y += 1;
      }
      cout << ans << "\n";
    }
  }
  int main() {
    judge();
    fastio();
    pre();
    int t = 1;
    for (int i = 0; i < t; i++) {
      solve2();
      cout << " \n";
    }
    cerr << "Time : " << ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
    return 0;
  }
  