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;
  template <typename T>
  struct compF {
    bool operator()(const pair<T, T> &a, const pair<T, T> &b) {
      return a.first < b.first;
    }
  };
  template <typename T>
  struct compS {
    bool operator()(const pair<T, T> &a, const pair<T, T> &b) {
      return a.second < b.second;
    }
  };
  template <typename T>
  void printMatrix(const string &txt, vector<vector<T>> &a) {
    cout << txt;
    for (int i = 0; i < a.size(); ++i) {
      for (int j = 0; j < a[i].size(); ++j) cout << a[i][j] << ' ';
      cout << endl;
    }
  }
  template <typename T>
  void printVector(const string &txt, vector<T> &a) {
    cout << txt;
    for (int i = 0; i < a.size(); ++i) cout << a[i] << ' ';
    cout << endl;
  }
  template <typename T>
  void readVector(int n, vector<T> &a) {
    if (a.size() < n) a.resize(n);
    for (int i = 0; i < n; ++i) cin >> a[i];
  }
  template <typename T>
  T lb(const set<T> &a, T x) {
    auto it = a.lower_bound(x);
    if (it == a.end()) return -1;
    return *it;
  }
  template <typename T>
  T ub(const set<T> &a, T x) {
    auto it = a.upper_bound(x);
    if (it == a.end()) return -1;
    return *it;
  }
  template <typename T>
  T rlb(const set<T> &a, T x) {
    auto it = a.lower_bound(x);
    if (it == a.begin()) {
      if (*it > x) return -1;
      return *it;
    }
    it = prev(it);
    return *it;
  }
  template <typename T>
  T rub(const set<T> &a, T x) {
    auto it = a.lower_bound(x);
    if (it == a.begin()) {
      return -1;
    }
    it = prev(it);
    return *it;
  }
  template <typename T1, typename T2>
  bool exist(map<T1, T2> &m, T1 x) {
    if (m.find(x) == m.end()) return false;
    return true;
  }
  long long gcd(long long a, long long b) {
    while (a != 0 && b != 0) {
      if (b > a)
        b %= a;
      else
        a %= b;
    }
    return a + b;
  }
  long long minFactor(long long x) {
    for (long long i = 2; i * i <= x; ++i)
      if (x % i == 0) return i;
    return x;
  }
  long long allFactors(long long x, vector<long long> &prms,
                       vector<long long> &prm_cnt) {
    prms.clear();
    prm_cnt.clear();
    while (x > 1) {
      long long t = minFactor(x);
      prms.push_back(t);
      int s = prms.size() - 1;
      prm_cnt.push_back(0);
      while (x % t == 0) {
        x /= t;
        ++prm_cnt[s];
      }
    }
    return 0;
  }
  long long add(long long a, long long b) {
    if (a > 998244353) a %= 998244353;
    if (b > 998244353) b %= 998244353;
    return (a + b) % 998244353;
  }
  long long sub(long long a, long long b) {
    if (a > 998244353) a %= 998244353;
    if (b > 998244353) b %= 998244353;
    long long x = a - b;
    while (x < 0) x += 998244353;
    return x;
  }
  long long mul(long long a, long long b) {
    if (a > 998244353) a %= 998244353;
    if (b > 998244353) b %= 998244353;
    return (a * b) % 998244353;
  }
  long long pow_(long long a, long long b) {
    if (b == 0) return 1ll;
    if (b % 2 == 0) {
      long long k = pow_(a, b / 2);
      return mul(k, k);
    } else {
      long long k = pow_(a, b - 1);
      return mul(a, k);
    }
  }
  int a[1000][1000] = {0, 0};
  int n, m, q;
  long long dp[1000][1000][2] = {0, 0, 0};
  pair<int, int> st1(int x, int y) {
    int len1 = 0;
    int xx = x, yy = y;
    while (xx >= 0 && yy >= 0 && !a[xx][yy]) {
      len1++;
      if (len1 % 2 == 1)
        --xx;
      else
        --yy;
    }
    xx = x, yy = y;
    int len2 = 0;
    while (xx < n && yy < m && !a[xx][yy]) {
      len2++;
      if (len2 % 2 == 1)
        ++yy;
      else
        ++xx;
    }
    return {len1, len2};
  }
  pair<int, int> st2(int x, int y) {
    int len1 = 0;
    int xx = x, yy = y;
    while (xx >= 0 && yy >= 0 && !a[xx][yy]) {
      len1++;
      if (len1 % 2 == 1)
        --yy;
      else
        --xx;
    }
    xx = x, yy = y;
    int len2 = 0;
    while (xx < n && yy < m && !a[xx][yy]) {
      len2++;
      if (len2 % 2 == 1)
        ++xx;
      else
        ++yy;
    }
    return {len1, len2};
  }
  long long C(long long x) { return (x * (x - 1)) / 2; }
  long long R(long long x) { return (x * (x + 1)) / 2; }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie();
    cin >> n >> m >> q;
    unsigned long long curr_res = 0;
    dp[0][0][0] = dp[0][0][1] = 0;
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < m; ++j) {
        if (i == 0 && j == 0) continue;
        if (j > 0) dp[i][j][0] += dp[i][j - 1][1] + 1;
        if (i > 0) dp[i][j][1] += dp[i - 1][j][0] + 1;
      }
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < m; ++j)
        for (int k = 0; k < 2; ++k) {
          curr_res += max(dp[i][j][k], 0ll);
        }
    curr_res += n * m;
    while (q--) {
      int x, y;
      cin >> x >> y;
      --x;
      --y;
      if (a[x][y] == 1) {
        a[x][y] = 0;
        pair<int, int> p1 = st1(x, y);
        pair<int, int> p2 = st2(x, y);
        int l1 = p1.first - 1, l2 = p1.second - 1;
        curr_res += l1;
        curr_res += l2;
        curr_res += l1 * l2;
        l1 = p2.first - 1, l2 = p2.second - 1;
        curr_res += l1;
        curr_res += l2;
        curr_res += l1 * l2;
        ++curr_res;
      } else {
        pair<int, int> p1 = st1(x, y);
        pair<int, int> p2 = st2(x, y);
        int l1 = p1.first - 1, l2 = p1.second - 1;
        curr_res -= l1;
        curr_res -= l2;
        curr_res -= l1 * l2;
        l1 = p2.first - 1, l2 = p2.second - 1;
        curr_res -= l1;
        curr_res -= l2;
        curr_res -= l1 * l2;
        curr_res--;
        a[x][y] = 1;
      }
      cout << curr_res << endl;
    }
    return 0;
  }
  