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;
  string to_string(string s) { return '"' + s + '"'; }
  string to_string(char s) { return string(1, s); }
  string to_string(const char *s) { return to_string((string)s); }
  string to_string(bool b) { return (b ? "true" : "false"); }
  template <typename A>
  string to_string(A);
  template <typename A, typename B>
  string to_string(pair<A, B> p) {
    return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
  }
  template <typename A>
  string to_string(A v) {
    bool f = 1;
    string r = "{";
    for (const auto &x : v) {
      if (!f) r += ", ";
      f = 0;
      r += to_string(x);
    }
    return r + "}";
  }
  void debug_out() { cout << endl; }
  void show() { cout << endl; }
  void pret() {
    cout << endl;
    exit(0);
  }
  template <typename Head, typename... Tail>
  void debug_out(Head H, Tail... T) {
    cout << " " << to_string(H);
    debug_out(T...);
  }
  template <typename Head, typename... Tail>
  void show(Head H, Tail... T) {
    cout << H << " ";
    show(T...);
  }
  template <typename Head, typename... Tail>
  void pret(Head H, Tail... T) {
    cout << H << " ";
    pret(T...);
  }
  const long long MOD = 1e9 + 7;
  const long long maxn = 1000 + 10;
  long long n, m, q;
  bool a[maxn][maxn];
  long long dp[maxn][maxn][2];
  long long find(long long x, long long y, long long t) {
    long long &ans = dp[x][y][t];
    if (ans != -1) return ans;
    ans = 0;
    if (t == 0) {
      if (x + 1 > n) {
        ans = 0;
      } else
        ans = 1 + find(x + 1, y, 1 - t);
    } else {
      if (y + 1 > m)
        ans = 0;
      else
        ans = 1 + find(x, y + 1, 1 - t);
    }
    return ans;
  }
  long long leftt(long long x, long long y) {
    long long i = x, j = y;
    bool flag = 0;
    long long curr = 0, temp = 0;
    while (i >= 1 and j >= 1) {
      if (a[i][j]) break;
      curr++;
      if (!flag)
        j--;
      else
        i--;
      flag = 1 - flag;
    }
    flag = 0;
    i = x, j = y;
    while (i <= n and j <= m) {
      if (a[i][j]) break;
      temp++;
      if (!flag)
        i++;
      else
        j++;
      flag = 1 - flag;
    }
    return curr * temp;
  }
  long long upp(long long x, long long y) {
    long long i = x, j = y;
    bool flag = 0;
    long long curr = 0, temp = 0;
    while (i >= 1 and j >= 1) {
      if (a[i][j]) break;
      curr++;
      if (!flag)
        i--;
      else
        j--;
      flag = 1 - flag;
    }
    flag = 0;
    i = x, j = y;
    while (i <= n and j <= m) {
      if (a[i][j]) break;
      temp++;
      if (!flag)
        j++;
      else
        i++;
      flag = 1 - flag;
    }
    return curr * temp;
  }
  int32_t main() {
    cin.tie(0);
    ios_base::sync_with_stdio(0);
    ;
    cin >> n >> m >> q;
    long long ans = 0;
    for (long long i = 1; i <= (long long)n; i++) {
      for (long long j = 1; j <= (long long)m; j++) {
        dp[i][j][0] = -1;
        dp[i][j][1] = -1;
      }
    }
    for (long long i = 1; i <= (long long)n; i++) {
      for (long long j = 1; j <= (long long)m; j++) {
        ans += find(i, j, 0ll);
        ans += find(i, j, 1ll);
        ans += 1;
      }
    }
    while (q--) {
      long long x, y;
      cin >> x >> y;
      long long flag = a[x][y];
      a[x][y] = 0;
      long long curr = leftt(x, y) + upp(x, y) - 1;
      if (flag) {
        ans += curr;
      } else {
        ans -= curr;
      }
      a[x][y] = 1 - flag;
      show(ans);
    }
    return 0;
  }
  