#include <bits/stdc++.h>
using namespace std;
int n, m, q;
int mmp[1010][1010];
long long ans = 0;
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m >> q;
  long long A = min(m, n), D = (m - n > 0) ? (m - n) : (n - m);
  ans = 2ll * (A - 1ll) * (A) * (2ll * A - 1ll) / 3ll + A * (A - 1ll) +
        D * A * (2ll * A - 1ll) + 1ll * n * m;
  int x, y;
  for (int i = 1; i <= q; i++) {
    cin >> x >> y;
    int xt = x, yt = y;
    long long up = 0, down = 0;
    for (int j = 1; j > -1; j++) {
      if (((j & 1) != 0) && mmp[xt - 1][yt] == 0 && xt - 1 >= 1)
        xt--;
      else if (((j & 1) == 0) && mmp[xt][yt - 1] == 0 && yt - 1 >= 1)
        yt--;
      else
        break;
      up++;
    }
    xt = x, yt = y;
    for (int j = 0; j >= -1; j++) {
      if (((j & 1) != 0) && mmp[xt + 1][yt] == 0 && xt + 1 <= n)
        xt++;
      else if (((j & 1) == 0) && mmp[xt][yt + 1] == 0 && yt + 1 <= m)
        yt++;
      else
        break;
      down++;
    }
    if (mmp[x][y])
      ans += (1ll * up * down + up + down + 1);
    else
      ans -= (1ll * up * down + up + down + 1);
    xt = x, yt = y;
    up = 0, down = 0;
    for (int j = 0; j > -1; j++) {
      if (((j & 1) != 0) && mmp[xt - 1][yt] == 0 && xt - 1 >= 1)
        xt--;
      else if (((j & 1) == 0) && mmp[xt][yt - 1] == 0 && yt - 1 >= 1)
        yt--;
      else
        break;
      up++;
    }
    xt = x, yt = y;
    for (int j = 1; j > -1; j++) {
      if (((j & 1) != 0) && mmp[xt + 1][yt] == 0 && xt + 1 <= n)
        xt++;
      else if (((j & 1) == 0) && mmp[xt][yt + 1] == 0 && yt + 1 <= m)
        yt++;
      else
        break;
      down++;
    }
    if (mmp[x][y])
      ans += (1ll * up * down + up + down + 1);
    else
      ans -= (1ll * up * down + up + down + 1);
    if (mmp[x][y] == 0)
      ans++;
    else
      ans--;
    mmp[x][y] ^= 1;
    cout << ans << endl;
  }
  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;
int n, m, q;
int mmp[1010][1010];
long long ans = 0;
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m >> q;
  long long A = min(m, n), D = (m - n > 0) ? (m - n) : (n - m);
  ans = 2ll * (A - 1ll) * (A) * (2ll * A - 1ll) / 3ll + A * (A - 1ll) +
        D * A * (2ll * A - 1ll) + 1ll * n * m;
  int x, y;
  for (int i = 1; i <= q; i++) {
    cin >> x >> y;
    int xt = x, yt = y;
    long long up = 0, down = 0;
    for (int j = 1; j > -1; j++) {
      if (((j & 1) != 0) && mmp[xt - 1][yt] == 0 && xt - 1 >= 1)
        xt--;
      else if (((j & 1) == 0) && mmp[xt][yt - 1] == 0 && yt - 1 >= 1)
        yt--;
      else
        break;
      up++;
    }
    xt = x, yt = y;
    for (int j = 0; j >= -1; j++) {
      if (((j & 1) != 0) && mmp[xt + 1][yt] == 0 && xt + 1 <= n)
        xt++;
      else if (((j & 1) == 0) && mmp[xt][yt + 1] == 0 && yt + 1 <= m)
        yt++;
      else
        break;
      down++;
    }
    if (mmp[x][y])
      ans += (1ll * up * down + up + down + 1);
    else
      ans -= (1ll * up * down + up + down + 1);
    xt = x, yt = y;
    up = 0, down = 0;
    for (int j = 0; j > -1; j++) {
      if (((j & 1) != 0) && mmp[xt - 1][yt] == 0 && xt - 1 >= 1)
        xt--;
      else if (((j & 1) == 0) && mmp[xt][yt - 1] == 0 && yt - 1 >= 1)
        yt--;
      else
        break;
      up++;
    }
    xt = x, yt = y;
    for (int j = 1; j > -1; j++) {
      if (((j & 1) != 0) && mmp[xt + 1][yt] == 0 && xt + 1 <= n)
        xt++;
      else if (((j & 1) == 0) && mmp[xt][yt + 1] == 0 && yt + 1 <= m)
        yt++;
      else
        break;
      down++;
    }
    if (mmp[x][y])
      ans += (1ll * up * down + up + down + 1);
    else
      ans -= (1ll * up * down + up + down + 1);
    if (mmp[x][y] == 0)
      ans++;
    else
      ans--;
    mmp[x][y] ^= 1;
    cout << ans << endl;
  }
  return 0;
}

```