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;
  template <class T, class U>
  inline bool chmax(T& a, U b) {
    if (a < b) {
      a = b;
      return true;
    }
    return false;
  }
  template <class T, class U>
  inline bool chmin(T& a, U b) {
    if (a > b) {
      a = b;
      return true;
    }
    return false;
  }
  constexpr int INF = 1000000000;
  constexpr ll llINF = 3000000000000000000;
  constexpr int mod = 998244353;
  constexpr double eps = 1e-10;
  const double pi = acos(-1);
  int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};
  int Random(int mi, int ma) {
    random_device rnd;
    mt19937 mt(rnd());
    uniform_int_distribution<int> engine(mi, ma);
    return engine(mt);
  }
  bool prime(int a) {
    if (a == 1) return false;
    for (int i = 2; i * i <= a; i++) {
      if (a % i == 0) return false;
    }
    return true;
  }
  vector<int> primes;
  void init_prime(int n) {
    primes.push_back(2);
    for (int i = 3; i <= n; i += 2) {
      bool f = true;
      for (int j : primes) {
        if (j * j > i) break;
        if (i % j == 0) {
          f = false;
          break;
        }
      }
      if (f) primes.push_back(i);
    }
  }
  ll modpow(ll a, ll b, ll m = mod) {
    ll res = 1;
    while (b) {
      if (b & 1) {
        res *= a;
        res %= m;
      }
      a *= a;
      a %= m;
      b >>= 1;
    }
    return res;
  }
  vector<ll> inv, fact, factinv;
  void init_fact(int n) {
    inv.resize(n + 1);
    fact.resize(n + 1);
    factinv.resize(n + 1);
    inv[0] = 0;
    inv[1] = 1;
    fact[0] = 1;
    factinv[0] = 1;
    for (ll i = 1; i <= n; i++) {
      if (i >= 2) inv[i] = mod - ((mod / i) * inv[mod % i] % mod);
      fact[i] = (fact[i - 1] * i) % mod;
      factinv[i] = factinv[i - 1] * inv[i] % mod;
    }
  }
  ll _inv(ll a, ll m = mod) {
    ll b = m, u = 1, v = 0;
    while (b) {
      ll t = a / b;
      a -= t * b;
      swap(a, b);
      u -= t * v;
      swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
  }
  ll comb(int a, int b) {
    if (a < b) return 0;
    if (a < 0) return 0;
    return fact[a] * factinv[a - b] % mod * factinv[b] % mod;
  }
  bool flag[1010][1010];
  int a[1010][1010];
  ll ans = 0;
  void solve() {
    int n, m, q;
    cin >> n >> m >> q;
    for (int i = n - 1; i >= 0; i--) {
      for (int j = m - 1; j >= 0; j--) {
        a[i][j] = (i + 1 < n && j + 1 < m ? a[i + 1][j + 1] : 0) + 1;
      }
    }
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        if (j + 1 < m)
          ans += min(a[i][j], a[i][j + 1]) * 2 + (a[i][j] > a[i][j + 1]) - 1;
        if (i + 1 < n)
          ans += min(a[i][j], a[i + 1][j]) * 2 + (a[i][j] > a[i + 1][j]) - 1;
      }
    }
    ans += n * m;
    while (q--) {
      int x, y;
      cin >> x >> y;
      x--, y--;
      if (flag[x][y]) {
        ans++;
        flag[x][y] = false;
        for (int i = x, j = y; i >= 0 && j >= 0 && !flag[i][j]; i--, j--) {
          if (i < x) {
            if (j + 1 < m)
              ans -= min(a[i][j], a[i][j + 1]) * 2 + (a[i][j] > a[i][j + 1]) - 1;
            if (i + 1 < n)
              ans -= min(a[i][j], a[i + 1][j]) * 2 + (a[i][j] > a[i + 1][j]) - 1;
            if (j && !flag[i][j - 1])
              ans -= min(a[i][j - 1], a[i][j]) * 2 + (a[i][j - 1] > a[i][j]) - 1;
            if (i && !flag[i - 1][j])
              ans -= min(a[i - 1][j], a[i][j]) * 2 + (a[i - 1][j] > a[i][j]) - 1;
          }
          a[i][j] = (i + 1 < n && j + 1 < m ? a[i + 1][j + 1] : 0) + 1;
          if (j + 1 < m)
            ans += min(a[i][j], a[i][j + 1]) * 2 + (a[i][j] > a[i][j + 1]) - 1;
          if (i + 1 < n)
            ans += min(a[i][j], a[i + 1][j]) * 2 + (a[i][j] > a[i + 1][j]) - 1;
          if (j && !flag[i][j - 1])
            ans += min(a[i][j - 1], a[i][j]) * 2 + (a[i][j - 1] > a[i][j]) - 1;
          if (i && !flag[i - 1][j])
            ans += min(a[i - 1][j], a[i][j]) * 2 + (a[i - 1][j] > a[i][j]) - 1;
        }
      } else {
        ans--;
        for (int i = x, j = y; i >= 0 && j >= 0 && !flag[i][j]; i--, j--) {
          if (j + 1 < m)
            ans -= min(a[i][j], a[i][j + 1]) * 2 + (a[i][j] > a[i][j + 1]) - 1;
          if (i + 1 < n)
            ans -= min(a[i][j], a[i + 1][j]) * 2 + (a[i][j] > a[i + 1][j]) - 1;
          if (j && !flag[i][j - 1])
            ans -= min(a[i][j - 1], a[i][j]) * 2 + (a[i][j - 1] > a[i][j]) - 1;
          if (i && !flag[i - 1][j])
            ans -= min(a[i - 1][j], a[i][j]) * 2 + (a[i - 1][j] > a[i][j]) - 1;
          a[i][j] = (i == x ? 0 : a[i + 1][j + 1] + 1);
          if (i < x) {
            if (j + 1 < m)
              ans += min(a[i][j], a[i][j + 1]) * 2 + (a[i][j] > a[i][j + 1]) - 1;
            if (i + 1 < n)
              ans += min(a[i][j], a[i + 1][j]) * 2 + (a[i][j] > a[i + 1][j]) - 1;
            if (j && !flag[i][j - 1])
              ans += min(a[i][j - 1], a[i][j]) * 2 + (a[i][j - 1] > a[i][j]) - 1;
            if (i && !flag[i - 1][j])
              ans += min(a[i - 1][j], a[i][j]) * 2 + (a[i - 1][j] > a[i][j]) - 1;
          }
        }
        flag[x][y] = true;
      }
      cout << ans << endl;
    }
  }
  int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    solve();
  }
  