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 ld = long double;
  using ull = unsigned long long;
  using uint = unsigned;
  inline ll popcnt(ull a) { return __builtin_popcountll(a); }
  ll intpow(ll a, ll b) {
    ll ans = 1;
    while (b) {
      if (b & 1) ans *= a;
      a *= a;
      b /= 2;
    }
    return ans;
  }
  ll intpow(ll a, ll b, ll m) {
    ll ans = 1;
    while (b) {
      if (b & 1) (ans *= a) %= m;
      (a *= a) %= m;
      b /= 2;
    }
    return ans;
  }
  template <class T>
  inline bool chmax(T& a, T b) {
    if (a < b) {
      a = b;
      return 1;
    }
    return 0;
  }
  template <class T>
  inline bool chmin(T& a, T b) {
    if (a > b) {
      a = b;
      return 1;
    }
    return 0;
  }
  template <class T>
  using vc = vector<T>;
  template <class T>
  using vvc = vector<vc<T>>;
  template <class T>
  using vvvc = vector<vvc<T>>;
  template <class T>
  using vvvvc = vector<vvvc<T>>;
  template <class T>
  using pq = priority_queue<T>;
  template <class T>
  using pqg = priority_queue<T, vector<T>, greater<T>>;
  template <class T>
  void scan(T& a) {
    cin >> a;
  }
  template <class T>
  void scan(vector<T>& a) {
    for (auto&& i : a) scan(i);
  }
  void in() {}
  template <class Head, class... Tail>
  void in(Head& head, Tail&... tail) {
    scan(head);
    in(tail...);
  }
  void print() { cout << ' '; }
  template <class T>
  void print(const T& a) {
    cout << a;
  }
  template <class T>
  void print(const vector<T>& a) {
    if (a.empty()) return;
    print(a[0]);
    for (auto i = a.begin(); ++i != a.end();) {
      cout << ' ';
      print(*i);
    }
  }
  int out() {
    cout << '\n';
    return 0;
  }
  template <class T>
  int out(const T& t) {
    print(t);
    cout << '\n';
    return 0;
  }
  template <class Head, class... Tail>
  int out(const Head& head, const Tail&... tail) {
    print(head);
    cout << ' ';
    out(tail...);
    return 0;
  }
  set<int> se[3030];
  const int offset = 1010;
  void solve() {
    int n, m, q;
    in(n, m, q);
    bool check = 0;
    if (n > m) {
      check = 1;
      swap(n, m);
    }
    vvc<int> lock(n, vc<int>(m, 0));
    auto f = [&](int x) -> int { return x + offset; };
    for (int k = -m + 1; k <= n - 1; ++k) {
      se[f(k)].insert(-1);
      se[f(k)].insert(n);
      if (k > 0) {
        se[f(k)].insert(k - 1);
      }
      if (k < n - m) {
        se[f(k)].insert(k + m);
      }
    }
    auto ok = [&](int x) -> bool { return -m + 1 <= x and x <= n - 1; };
    ll sum = 0;
    vvvc<ll> dp(n, vvc<ll>(m, vc<ll>(3, 0)));
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < m; ++j) {
        dp[i][j][2] += 1;
        if (i - 1 >= 0) dp[i][j][0] += dp[i - 1][j][1] + dp[i - 1][j][2];
        if (j - 1 >= 0) dp[i][j][1] += dp[i][j - 1][0] + dp[i][j - 1][2];
        for (int k = 0; k < 3; ++k) sum += dp[i][j][k];
      }
    }
    for (int i = 0; i < q; ++i) {
      int x, y;
      in(x, y);
      --x, --y;
      if (check) swap(x, y);
      ll tmp = 0;
      if (lock[x][y]) {
        se[f(x - y)].erase(x);
      }
      for (int k = -1; k <= 0; ++k) {
        if (!(ok(x - y + k) and ok(x - y + k + 1))) {
          tmp++;
          continue;
        }
        auto ite = se[f(x - y + k)].upper_bound(x);
        auto ite2 = se[f(x - y + k + 1)].upper_bound(x);
        auto ite3 = se[f(x - y + k)].lower_bound(x);
        --ite3;
        auto ite4 = se[f(x - y + k + 1)].lower_bound(x);
        --ite4;
        ll left = 0, right = 0;
        ;
        if (*ite >= *ite2) {
          right = (*ite2 - x - 1) * 2;
          if (k == -1) ++right;
        } else {
          right = (*ite - x - 1) * 2 + 1;
          if (k == -1) ++right;
        }
        if (*ite3 >= *ite4) {
          left = (x - *ite3 - 1) * 2;
          if (k == 0) ++left;
        } else {
          left = (x - *ite4 - 1) * 2 + 1;
          if (k == 0) ++left;
        }
        if (left < 0) left = 0;
        if (right < 0) right = 0;
        if (k == -1 and se[f(x - y + k)].find(x) != se[f(x - y + k)].end())
          right = 0;
        if (k == 0 and se[f(x - y + k + 1)].find(x) != se[f(x - y + k + 1)].end())
          left = 0;
        ;
        tmp += (left + 1) * (right + 1);
      }
      tmp -= 1;
      ;
      ;
      if (lock[x][y]) {
        sum += tmp;
      } else {
        se[f(x - y)].insert(x);
        sum -= tmp;
      }
      lock[x][y] ^= 1;
      out(sum);
    }
  }
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    solve();
    return 0;
  }
  