#include <bits/stdc++.h>
using namespace std;
inline long long md(long long m, long long v) { return ((v % m) + m) % m; }
void solve() {
  int Y, X, Q;
  cin >> Y >> X >> Q;
  long long ans = 0;
  for (int y = 0; y < Y; y++) {
    for (int x = 0; x < X; x++) {
      int x_space = X - 1 - x;
      int y_space = Y - 1 - y;
      ans++;
      {
        int R = min(x_space, y_space + 1);
        int D = min(y_space, R);
        ans += R + D;
      }
      {
        int D = min(y_space, x_space + 1);
        int R = min(x_space, D);
        ans += R + D;
      }
    }
  }
  vector<vector<int> > lock(Y, vector<int>(X, 0));
  auto fn = [&](int y, int x, int dy, int dx) -> int {
    int ret = 0;
    while (1) {
      y += dy, x += dx;
      if (y < 0 || x < 0 || y >= Y || x >= X || lock[y][x]) break;
      ret++;
      swap(dy, dx);
    }
    return ret;
  };
  while (Q--) {
    int y, x;
    cin >> y >> x;
    y--, x--;
    bool locked = lock[y][x];
    {
      int up = fn(y, x, -1, 0);
      int down = fn(y, x, 0, 1);
      ans += (locked ? 1 : -1) * ((up + 1) * (down + 1) - 1);
    }
    {
      int up = fn(y, x, 0, -1);
      int down = fn(y, x, 1, 0);
      ans += (locked ? 1 : -1) * ((up + 1) * (down + 1) - 1);
    }
    ans += (locked ? 1 : -1);
    lock[y][x] ^= 1;
    cout << ans << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  solve();
  return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
inline long long md(long long m, long long v) { return ((v % m) + m) % m; }
void solve() {
  int Y, X, Q;
  cin >> Y >> X >> Q;
  long long ans = 0;
  for (int y = 0; y < Y; y++) {
    for (int x = 0; x < X; x++) {
      int x_space = X - 1 - x;
      int y_space = Y - 1 - y;
      ans++;
      {
        int R = min(x_space, y_space + 1);
        int D = min(y_space, R);
        ans += R + D;
      }
      {
        int D = min(y_space, x_space + 1);
        int R = min(x_space, D);
        ans += R + D;
      }
    }
  }
  vector<vector<int> > lock(Y, vector<int>(X, 0));
  auto fn = [&](int y, int x, int dy, int dx) -> int {
    int ret = 0;
    while (1) {
      y += dy, x += dx;
      if (y < 0 || x < 0 || y >= Y || x >= X || lock[y][x]) break;
      ret++;
      swap(dy, dx);
    }
    return ret;
  };
  while (Q--) {
    int y, x;
    cin >> y >> x;
    y--, x--;
    bool locked = lock[y][x];
    {
      int up = fn(y, x, -1, 0);
      int down = fn(y, x, 0, 1);
      ans += (locked ? 1 : -1) * ((up + 1) * (down + 1) - 1);
    }
    {
      int up = fn(y, x, 0, -1);
      int down = fn(y, x, 1, 0);
      ans += (locked ? 1 : -1) * ((up + 1) * (down + 1) - 1);
    }
    ans += (locked ? 1 : -1);
    lock[y][x] ^= 1;
    cout << ans << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  solve();
  return 0;
}

```