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 <class T>
  inline void chmin(T &a, const T &val) {
    if (a > val) a = val;
  }
  template <class T>
  inline void chmax(T &a, const T &val) {
    if (a < val) a = val;
  }
  template <class T>
  long long sqr(T a) {
    return 1ll * a * a;
  }
  template <class T>
  void compress(T &a) {
    sort(a.begin(), a.end());
    a.resize(unique(a.begin(), a.end()) - a.begin());
  }
  const double pi = acos(-1.00);
  const double eps = 1e-6;
  const int INF = 2e9;
  const int MOD = 1e9 + 7;
  const int dx[4] = {0, 0, -1, 1};
  const int dy[4] = {-1, 1, 0, 0};
  const int N = 1e3 + 1;
  struct pack {
    int x, y, k;
  };
  vector<pack> a[N];
  vector<pack> b[N];
  vector<int> posA[N][N], posB[N][N];
  long long sumA[N], sumB[N];
  bool mark[N][N];
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n, m, q;
    cin >> n >> m >> q;
    for (int j = 1; j <= m; j++) {
      int cur = j;
      for (int i = 1; i <= n; i++) {
        a[j].push_back({i, cur, 0});
        posA[i][cur].push_back(j);
        cur++;
        if (cur > m) break;
        a[j].push_back({i, cur, 0});
        posA[i][cur].push_back(j);
      }
    }
    for (int i = 1; i <= n; i++) {
      int cur = i;
      for (int j = 1; j <= m; j++) {
        b[i].push_back({cur, j, 0});
        posB[cur][j].push_back(i);
        cur++;
        if (cur > n) break;
        b[i].push_back({cur, j, 0});
        posB[cur][j].push_back(i);
      }
    }
    long long tot = n * m;
    for (int i = 1; i <= m; i++) {
      int sz = a[i].size();
      tot += 1ll * sz * (sz - 1) / 2;
      sumA[i] = 1ll * sz * (sz - 1) / 2;
    }
    for (int i = 1; i <= n; i++) {
      int sz = b[i].size();
      tot += 1ll * sz * (sz - 1) / 2;
      sumB[i] = 1ll * sz * (sz - 1) / 2;
    }
    auto updateA = [&](int x, int y, int id) {
      if (id == 0) return;
      for (pack &cur : a[id]) {
        if (cur.x == x && cur.y == y) {
          cur.k = 1 - cur.k;
        }
      }
      tot -= sumA[id];
      int cnt = 0;
      long long tmp = 0;
      for (pack &cur : a[id]) {
        if (cur.k == 1) {
          tmp += 1ll * cnt * (cnt - 1) / 2;
          cnt = 0;
        } else
          cnt++;
      }
      tmp += 1ll * cnt * (cnt - 1) / 2;
      sumA[id] = tmp;
      tot += tmp;
    };
    auto updateB = [&](int x, int y, int id) {
      for (pack &cur : b[id]) {
        if (cur.x == x && cur.y == y) {
          cur.k = 1 - cur.k;
        }
      }
      tot -= sumB[id];
      int cnt = 0;
      long long tmp = 0;
      for (pack &cur : b[id]) {
        if (cur.k == 1) {
          tmp += 1ll * cnt * (cnt - 1) / 2;
          cnt = 0;
        } else
          cnt++;
      }
      tmp += 1ll * cnt * (cnt - 1) / 2;
      sumB[id] = tmp;
      tot += tmp;
    };
    while (q--) {
      int x, y;
      cin >> x >> y;
      if (mark[x][y])
        tot++;
      else
        tot--;
      mark[x][y] = (!mark[x][y]);
      for (int id : posA[x][y]) updateA(x, y, id);
      for (int id : posB[x][y]) updateB(x, y, id);
      cout << tot << '\n';
    }
    return 0;
  }
  