#include <bits/stdc++.h>
using namespace std;
const int N = 1003;
bool mp[N][N];
long long dp[N][N][2] = {0}, ans = 0;
int main() {
  int n, m, q, x, y, z;
  cin >> n >> m >> q;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (j > 1) dp[i][j][0] = dp[i][j - 1][1] + 1;
      if (i > 1) dp[i][j][1] = dp[i - 1][j][0] + 1;
      ans += dp[i][j][1] + dp[i][j][0] + 1;
    }
  }
  while (q--) {
    cin >> x >> y;
    mp[x][y] = !mp[x][y];
    z = min(n - x, m - y);
    if (mp[x][y]) {
      ans -= dp[x][y][1] + dp[x][y][0] + 1;
      dp[x][y][1] = dp[x][y][0] = 0;
    } else {
      if (y > 1) dp[x][y][0] = dp[x][y - 1][1] + !mp[x][y - 1];
      if (x > 1) dp[x][y][1] = dp[x - 1][y][0] + !mp[x - 1][y];
      ans += dp[x][y][1] + dp[x][y][0] + 1;
    }
    for (int i = 1;; i++) {
      if ((x + i <= n) && (y + i - 1 <= m) && !mp[x + i][y + i - 1]) {
        ans -= dp[x + i][y + i - 1][1];
        dp[x + i][y + i - 1][1] =
            dp[x + i - 1][y + i - 1][0] + !mp[x + i - 1][y + i - 1];
        ans += dp[x + i][y + i - 1][1];
      }
      if ((x + i - 1 <= n) && (y + i <= m) && !mp[x + i - 1][y + i]) {
        ans -= dp[i + x - 1][y + i][0];
        dp[x + i - 1][y + i][0] =
            dp[x + i - 1][y + i - 1][1] + !mp[x + i - 1][y + i - 1];
        ans += dp[i + x - 1][y + i][0];
      }
      if ((x + i <= n) && (y + i <= m) && !mp[x + i][y + i]) {
        ans -= dp[x + i][y + i][1] + dp[i + x][y + i][0];
        dp[x + i][y + i][0] = dp[x + i][y + i - 1][1] + !mp[x + i][y + i - 1];
        dp[x + i][y + i][1] = dp[x + i - 1][y + i][0] + !mp[x + i - 1][y + i];
        ans += dp[x + i][y + i][1] + dp[i + x][y + i][0];
      }
      if (i > z) break;
    }
    cout << ans << "\n";
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 = 1003;
bool mp[N][N];
long long dp[N][N][2] = {0}, ans = 0;
int main() {
  int n, m, q, x, y, z;
  cin >> n >> m >> q;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (j > 1) dp[i][j][0] = dp[i][j - 1][1] + 1;
      if (i > 1) dp[i][j][1] = dp[i - 1][j][0] + 1;
      ans += dp[i][j][1] + dp[i][j][0] + 1;
    }
  }
  while (q--) {
    cin >> x >> y;
    mp[x][y] = !mp[x][y];
    z = min(n - x, m - y);
    if (mp[x][y]) {
      ans -= dp[x][y][1] + dp[x][y][0] + 1;
      dp[x][y][1] = dp[x][y][0] = 0;
    } else {
      if (y > 1) dp[x][y][0] = dp[x][y - 1][1] + !mp[x][y - 1];
      if (x > 1) dp[x][y][1] = dp[x - 1][y][0] + !mp[x - 1][y];
      ans += dp[x][y][1] + dp[x][y][0] + 1;
    }
    for (int i = 1;; i++) {
      if ((x + i <= n) && (y + i - 1 <= m) && !mp[x + i][y + i - 1]) {
        ans -= dp[x + i][y + i - 1][1];
        dp[x + i][y + i - 1][1] =
            dp[x + i - 1][y + i - 1][0] + !mp[x + i - 1][y + i - 1];
        ans += dp[x + i][y + i - 1][1];
      }
      if ((x + i - 1 <= n) && (y + i <= m) && !mp[x + i - 1][y + i]) {
        ans -= dp[i + x - 1][y + i][0];
        dp[x + i - 1][y + i][0] =
            dp[x + i - 1][y + i - 1][1] + !mp[x + i - 1][y + i - 1];
        ans += dp[i + x - 1][y + i][0];
      }
      if ((x + i <= n) && (y + i <= m) && !mp[x + i][y + i]) {
        ans -= dp[x + i][y + i][1] + dp[i + x][y + i][0];
        dp[x + i][y + i][0] = dp[x + i][y + i - 1][1] + !mp[x + i][y + i - 1];
        dp[x + i][y + i][1] = dp[x + i - 1][y + i][0] + !mp[x + i - 1][y + i];
        ans += dp[x + i][y + i][1] + dp[i + x][y + i][0];
      }
      if (i > z) break;
    }
    cout << ans << "\n";
  }
  return 0;
}

```