#include <bits/stdc++.h>
using namespace std;
int n, m, a[1005][1005], f[1005][1005], ans;
int calc(int x, int y, int dx, int dy, int type) {
  int cnt = 0;
  if (type == 0) {
    while (x - dx >= 1 && y - dy >= 1) {
      x -= dx, y -= dy;
      if (!a[x][y]) break;
      ++cnt;
      if (dx == 1)
        dy = 1, dx = 0;
      else
        dy = 0, dx = 1;
    }
  } else {
    while (x + dx <= n && y + dy <= m) {
      x += dx, y += dy;
      if (!a[x][y]) break;
      ++cnt;
      if (dx == 1)
        dy = 1, dx = 0;
      else
        dy = 0, dx = 1;
    }
  }
  return cnt;
}
signed main() {
  int q;
  cin >> n >> m >> q;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) a[i][j] = 1;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j)
      ans += 1 + calc(i, j, 0, 1, 1) + calc(i, j, 1, 0, 1);
  while (q--) {
    int i, j;
    cin >> i >> j;
    int sum = (1 + calc(i, j, 0, 1, 1)) * (1 + calc(i, j, 1, 0, 0)) +
              (1 + calc(i, j, 1, 0, 1)) * (1 + calc(i, j, 0, 1, 0)) - 1;
    if (a[i][j])
      ans -= sum;
    else
      ans += sum;
    a[i][j] ^= 1;
    cout << ans << "\n";
  }
}

### Prompt

Generate 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;
int n, m, a[1005][1005], f[1005][1005], ans;
int calc(int x, int y, int dx, int dy, int type) {
  int cnt = 0;
  if (type == 0) {
    while (x - dx >= 1 && y - dy >= 1) {
      x -= dx, y -= dy;
      if (!a[x][y]) break;
      ++cnt;
      if (dx == 1)
        dy = 1, dx = 0;
      else
        dy = 0, dx = 1;
    }
  } else {
    while (x + dx <= n && y + dy <= m) {
      x += dx, y += dy;
      if (!a[x][y]) break;
      ++cnt;
      if (dx == 1)
        dy = 1, dx = 0;
      else
        dy = 0, dx = 1;
    }
  }
  return cnt;
}
signed main() {
  int q;
  cin >> n >> m >> q;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) a[i][j] = 1;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j)
      ans += 1 + calc(i, j, 0, 1, 1) + calc(i, j, 1, 0, 1);
  while (q--) {
    int i, j;
    cin >> i >> j;
    int sum = (1 + calc(i, j, 0, 1, 1)) * (1 + calc(i, j, 1, 0, 0)) +
              (1 + calc(i, j, 1, 0, 1)) * (1 + calc(i, j, 0, 1, 0)) - 1;
    if (a[i][j])
      ans -= sum;
    else
      ans += sum;
    a[i][j] ^= 1;
    cout << ans << "\n";
  }
}

```