#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
int n, m, Q;
long long ans, cnt, f[N][N][2];
bool vis[N][N];
void update(int i, int j) {
  if (j < 1 || j > m) return;
  ans -= (f[i][j][0] + f[i][j][1]);
  if (vis[i][j])
    f[i][j][0] = f[i][j][1] = 0;
  else
    f[i][j][0] = f[i][j + 1][1] + 1, f[i][j][1] = f[i + 1][j][0] + 1;
  ans += (f[i][j][0] + f[i][j][1]);
}
int main() {
  scanf("%d%d%d", &n, &m, &Q);
  for (int i = n; i >= 1; i--)
    for (int j = m; j >= 1; j--) update(i, j), cnt++;
  while (Q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    vis[x][y] ^= 1;
    if (vis[x][y] == 1)
      cnt--;
    else
      cnt++;
    for (int i = x, j = y - 1; i >= 1; i--, j--)
      update(i, j + 2), update(i, j + 1), update(i, j);
    cout << ans - cnt << "\n";
  }
  return 0;
}

### Prompt

Please formulate 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;
const int N = 1010;
int n, m, Q;
long long ans, cnt, f[N][N][2];
bool vis[N][N];
void update(int i, int j) {
  if (j < 1 || j > m) return;
  ans -= (f[i][j][0] + f[i][j][1]);
  if (vis[i][j])
    f[i][j][0] = f[i][j][1] = 0;
  else
    f[i][j][0] = f[i][j + 1][1] + 1, f[i][j][1] = f[i + 1][j][0] + 1;
  ans += (f[i][j][0] + f[i][j][1]);
}
int main() {
  scanf("%d%d%d", &n, &m, &Q);
  for (int i = n; i >= 1; i--)
    for (int j = m; j >= 1; j--) update(i, j), cnt++;
  while (Q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    vis[x][y] ^= 1;
    if (vis[x][y] == 1)
      cnt--;
    else
      cnt++;
    for (int i = x, j = y - 1; i >= 1; i--, j--)
      update(i, j + 2), update(i, j + 1), update(i, j);
    cout << ans - cnt << "\n";
  }
  return 0;
}

```