#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m, q;
  cin >> n >> m >> q;
  bool available[n][m];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      available[i][j] = true;
    }
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      for (int b = 0; b <= 1; b++) {
        int len = 1;
        while (true) {
          int x = i + (len + !b) / 2;
          int y = j + (len + b) / 2;
          if (x >= n || y >= m) {
            break;
          }
          len++;
        }
        ans += len - 1;
      }
    }
  }
  ans += n * m;
  while (q--) {
    int x, y;
    cin >> x >> y;
    x--, y--;
    vector<int> vec1;
    int new_ans1 = 1;
    for (int b : vector<int>({0, 1})) {
      int len = 1;
      while (true) {
        int x1 = x + (len + !b) / 2;
        int y1 = y + (len + b) / 2;
        if (x1 >= n || y1 >= m || !available[x1][y1]) {
          break;
        }
        len++;
      }
      new_ans1 += len;
      vec1.push_back(len);
    }
    vector<int> vec2;
    int new_ans2 = 1;
    for (int b = 0; b <= 1; b++) {
      int len = 1;
      while (true) {
        int x1 = x - (len + !b) / 2;
        int y1 = y - (len + b) / 2;
        if (x1 < 0 || y1 < 0 || !available[x1][y1]) {
          break;
        }
        len++;
      }
      new_ans2 += len;
      vec2.push_back(len);
    }
    int d = vec1[0] * vec2[1] + vec1[1] * vec2[0] - 1;
    if (available[x][y]) {
      ans -= d;
    } else {
      ans += d;
    }
    available[x][y] = !available[x][y];
    cout << ans << endl;
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
int main() {
  int n, m, q;
  cin >> n >> m >> q;
  bool available[n][m];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      available[i][j] = true;
    }
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      for (int b = 0; b <= 1; b++) {
        int len = 1;
        while (true) {
          int x = i + (len + !b) / 2;
          int y = j + (len + b) / 2;
          if (x >= n || y >= m) {
            break;
          }
          len++;
        }
        ans += len - 1;
      }
    }
  }
  ans += n * m;
  while (q--) {
    int x, y;
    cin >> x >> y;
    x--, y--;
    vector<int> vec1;
    int new_ans1 = 1;
    for (int b : vector<int>({0, 1})) {
      int len = 1;
      while (true) {
        int x1 = x + (len + !b) / 2;
        int y1 = y + (len + b) / 2;
        if (x1 >= n || y1 >= m || !available[x1][y1]) {
          break;
        }
        len++;
      }
      new_ans1 += len;
      vec1.push_back(len);
    }
    vector<int> vec2;
    int new_ans2 = 1;
    for (int b = 0; b <= 1; b++) {
      int len = 1;
      while (true) {
        int x1 = x - (len + !b) / 2;
        int y1 = y - (len + b) / 2;
        if (x1 < 0 || y1 < 0 || !available[x1][y1]) {
          break;
        }
        len++;
      }
      new_ans2 += len;
      vec2.push_back(len);
    }
    int d = vec1[0] * vec2[1] + vec1[1] * vec2[0] - 1;
    if (available[x][y]) {
      ans -= d;
    } else {
      ans += d;
    }
    available[x][y] = !available[x][y];
    cout << ans << endl;
  }
}

```