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;
  namespace IO_BUFF {
  const long long BS = (1 << 24) + 5;
  char Buffer[BS], *HD, *TL;
  inline long long gc() {
    if (HD == TL) TL = (HD = Buffer) + fread(Buffer, 1, BS, stdin);
    return (HD == TL) ? EOF : *HD++;
  }
  inline long long inn() {
    long long x, ch;
    while ((ch = gc()) < '0' || ch > '9')
      ;
    x = ch ^ '0';
    while ((ch = gc()) >= '0' && ch <= '9') x = (x << 1) + (x << 3) + (ch ^ '0');
    return x;
  }
  char ss[19999999], tt[20];
  long long ssl, ttl;
  inline long long print(long long x) {
    if (x < 0) ss[++ssl] = '-', x = (-x);
    if (!x) ss[++ssl] = '0';
    for (ttl = 0; x; x /= 10) tt[++ttl] = char(x % 10 + '0');
    for (; ttl; ttl--) ss[++ssl] = tt[ttl];
    return ss[++ssl] = '\n';
  }
  inline long long Flush() {
    return fwrite(ss + 1, sizeof(char), ssl, stdout), ssl = 0, 0;
  }
  long long read() {
    char c = getchar();
    long long x = 1;
    long long s = 0;
    while (c < '0' || c > '9') {
      if (c == '-') x = -1;
      c = getchar();
    }
    while (c >= '0' && c <= '9') {
      s = s * 10 + c - '0';
      c = getchar();
    }
    return s * x;
  }
  }  // namespace IO_BUFF
  using namespace IO_BUFF;
  long long T;
  const long long N = 1005;
  long long n, m, q, x, y, f[N][N][2], summ, cnt, vis[N][N], ma[N][N];
  struct node {
    long long x, y, id;
  };
  void solve() {
    n = read();
    m = read();
    q = read();
    cnt = summ = 0;
    for (long long i = 1; i <= n; i++) {
      for (long long k = 1; k <= m; k++) {
        if (i > 1) f[i][k][0] = 1;
        if (k > 1) f[i][k][1] = 1;
        if (i > 1) f[i][k][0] = (f[i][k][0] + f[i - 1][k][1]);
        if (k > 1) f[i][k][1] = (f[i][k][1] + f[i][k - 1][0]);
        summ = summ + f[i][k][0] + f[i][k][1];
        ma[i][k] = 0;
      }
    }
    cnt = n * m;
    for (long long i = 1; i <= q; i++) {
      x = read();
      y = read();
      if (ma[x][y]) {
        ma[x][y] = 0;
        if (x > 1 && !ma[x - 1][y]) f[x][y][0] = 1;
        if (y > 1 && !ma[x][y - 1]) f[x][y][1] = 1;
        if (x > 1) f[x][y][0] = (f[x][y][0] + f[x - 1][y][1]);
        if (y > 1) f[x][y][1] = (f[x][y][1] + f[x][y - 1][0]);
        summ = summ + f[x][y][0] + f[x][y][1];
        vector<node> v;
        for (long long i = 1; i <= min(n - x, m - y); i++) {
          v.push_back((node){x + i, y + i - 1, 0});
          v.push_back((node){x + i - 1, y + i, 1});
          v.push_back((node){x + i, y + i, 3});
        }
        if (n - x != m - y) {
          if (n - x > m - y) {
            v.push_back((node){x + (m - y) + 1, m, 0});
          } else {
            v.push_back((node){n, y + (n - x) + 1, 1});
          }
        }
        for (long long i = 0; i < v.size(); i++) {
          long long xx = v[i].x, yy = v[i].y;
          if (ma[xx][yy]) continue;
          summ = summ - f[xx][yy][0] - f[xx][yy][1];
          f[xx][yy][0] = f[xx][yy][1] = 0;
          if (xx > 1 && !ma[xx - 1][yy]) f[xx][yy][0] = 1;
          if (yy > 1 && !ma[xx][yy - 1]) f[xx][yy][1] = 1;
          if (xx > 1 && !ma[xx - 1][yy])
            f[xx][yy][0] = (f[xx][yy][0] + f[xx - 1][yy][1]);
          if (yy > 1 && !ma[xx][yy - 1])
            f[xx][yy][1] = (f[xx][yy][1] + f[xx][yy - 1][0]);
          summ = summ + f[xx][yy][0] + f[xx][yy][1];
        }
        cnt++;
      } else {
        ma[x][y] = 1;
        summ = summ - f[x][y][0] - f[x][y][1];
        f[x][y][0] = f[x][y][1] = 0;
        vector<node> v;
        for (long long i = 1; i <= min(n - x, m - y); i++) {
          v.push_back((node){x + i, y + i - 1, 0});
          v.push_back((node){x + i - 1, y + i, 1});
          v.push_back((node){x + i, y + i, 3});
        }
        if (n - x != m - y) {
          if (n - x > m - y) {
            v.push_back((node){x + (m - y) + 1, m, 0});
          } else {
            v.push_back((node){n, y + (n - x) + 1, 1});
          }
        }
        for (long long i = 0; i < v.size(); i++) {
          long long xx = v[i].x, yy = v[i].y;
          if (ma[xx][yy]) continue;
          summ = summ - f[xx][yy][0] - f[xx][yy][1];
          f[xx][yy][0] = f[xx][yy][1] = 0;
          if (xx > 1 && !ma[xx - 1][yy]) f[xx][yy][0] = 1;
          if (yy > 1 && !ma[xx][yy - 1]) f[xx][yy][1] = 1;
          if (xx > 1 && !ma[xx - 1][yy])
            f[xx][yy][0] = (f[xx][yy][0] + f[xx - 1][yy][1]);
          if (yy > 1 && !ma[xx][yy - 1])
            f[xx][yy][1] = (f[xx][yy][1] + f[xx][yy - 1][0]);
          summ = summ + f[xx][yy][0] + f[xx][yy][1];
        }
        cnt--;
      }
      cout << summ + cnt << '\n';
    }
    for (long long i = 1; i <= n; i++)
      for (long long k = 1; k <= n; k++) ma[i][k] = 0;
  }
  signed main() {
    T = 1;
    while (T--) {
      solve();
    }
    return 0;
  }
  