#include <bits/stdc++.h>
using namespace std;
int read() {
  char ch = getchar();
  int x = 1, s = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') x = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
  return x * s;
}
int M[1005][1005];
int main() {
  int n = read(), m = read(), q = read();
  int ans = n * m;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      int a = n - i, b = m - j;
      if (a > b)
        ans += b * 2 + 1;
      else
        ans += a * 2;
      if (b > a)
        ans += a * 2 + 1;
      else
        ans += b * 2;
    }
  }
  for (int i = 1; i <= q; i++) {
    int a = read(), b = read();
    int x = a, y = b, A = 0, B = 0, tmp = M[a][b];
    M[a][b] = 0;
    while (1) {
      if (x > 0 && y > 0 && M[x][y] == 0)
        A++, x--;
      else
        break;
      if (x > 0 && y > 0 && M[x][y] == 0)
        A++, y--;
      else
        break;
    }
    x = a, y = b;
    while (1) {
      if (x <= n && y <= m && M[x][y] == 0)
        B++, y++;
      else
        break;
      if (x <= n && y <= m && M[x][y] == 0)
        B++, x++;
      else
        break;
    }
    if (tmp)
      ans += A * B;
    else
      ans -= A * B;
    A = B = 0;
    x = a, y = b;
    while (1) {
      if (x > 0 && y > 0 && M[x][y] == 0)
        A++, y--;
      else
        break;
      if (x > 0 && y > 0 && M[x][y] == 0)
        A++, x--;
      else
        break;
    }
    x = a, y = b;
    while (1) {
      if (x <= n && y <= m && M[x][y] == 0)
        B++, x++;
      else
        break;
      if (x <= n && y <= m && M[x][y] == 0)
        B++, y++;
      else
        break;
    }
    if (tmp)
      ans += A * B - 1;
    else
      ans -= A * B - 1;
    printf("%d\n", ans);
    M[a][b] = tmp ^ 1;
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 read() {
  char ch = getchar();
  int x = 1, s = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') x = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
  return x * s;
}
int M[1005][1005];
int main() {
  int n = read(), m = read(), q = read();
  int ans = n * m;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      int a = n - i, b = m - j;
      if (a > b)
        ans += b * 2 + 1;
      else
        ans += a * 2;
      if (b > a)
        ans += a * 2 + 1;
      else
        ans += b * 2;
    }
  }
  for (int i = 1; i <= q; i++) {
    int a = read(), b = read();
    int x = a, y = b, A = 0, B = 0, tmp = M[a][b];
    M[a][b] = 0;
    while (1) {
      if (x > 0 && y > 0 && M[x][y] == 0)
        A++, x--;
      else
        break;
      if (x > 0 && y > 0 && M[x][y] == 0)
        A++, y--;
      else
        break;
    }
    x = a, y = b;
    while (1) {
      if (x <= n && y <= m && M[x][y] == 0)
        B++, y++;
      else
        break;
      if (x <= n && y <= m && M[x][y] == 0)
        B++, x++;
      else
        break;
    }
    if (tmp)
      ans += A * B;
    else
      ans -= A * B;
    A = B = 0;
    x = a, y = b;
    while (1) {
      if (x > 0 && y > 0 && M[x][y] == 0)
        A++, y--;
      else
        break;
      if (x > 0 && y > 0 && M[x][y] == 0)
        A++, x--;
      else
        break;
    }
    x = a, y = b;
    while (1) {
      if (x <= n && y <= m && M[x][y] == 0)
        B++, x++;
      else
        break;
      if (x <= n && y <= m && M[x][y] == 0)
        B++, y++;
      else
        break;
    }
    if (tmp)
      ans += A * B - 1;
    else
      ans -= A * B - 1;
    printf("%d\n", ans);
    M[a][b] = tmp ^ 1;
  }
  return 0;
}

```