#include <bits/stdc++.h>
using namespace std;
void fastIO() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
}
int N, M, Q;
bool arr[1100][1100];
int dp[1100][1100][2], ans;
void upd(int x, int y, int d) {
  if (x < 1 || y < 1) return;
  if (arr[x][y]) return;
  if (d == 0) {
    ans -= dp[x][y][d];
    dp[x][y][d] = dp[x][y + 1][1 - d];
    if (y + 1 <= M && !arr[x][y + 1]) dp[x][y][d]++;
    ans += dp[x][y][d];
    upd(x - 1, y, 1 - d);
  } else {
    ans -= dp[x][y][d];
    dp[x][y][d] = dp[x + 1][y][1 - d];
    if (x + 1 <= N && !arr[x + 1][y]) dp[x][y][d]++;
    ans += dp[x][y][d];
    upd(x, y - 1, 1 - d);
  }
}
int main() {
  fastIO();
  cin >> N >> M >> Q;
  for (int i = N; i >= 1; i--) {
    for (int j = M; j >= 1; j--) {
      if (j + 1 <= M) dp[i][j][0] = dp[i][j + 1][1] + 1;
      if (i + 1 <= N) dp[i][j][1] = dp[i + 1][j][0] + 1;
    }
  }
  ans = N * M;
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= M; j++) {
      ans += dp[i][j][0] + dp[i][j][1];
    }
  }
  for (int i = 1; i <= Q; i++) {
    int a, b;
    cin >> a >> b;
    if (arr[a][b] == false) {
      ans -= dp[a][b][0] + dp[a][b][1];
      dp[a][b][0] = 0;
      dp[a][b][1] = 0;
      ans--;
      arr[a][b] = true;
      upd(a, b - 1, 0);
      upd(a - 1, b, 1);
    } else {
      ans++;
      arr[a][b] = false;
      upd(a, b, 0);
      upd(a, b, 1);
    }
    cout << ans << "\n";
  }
}

### Prompt

Create a solution in cpp for 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;
void fastIO() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
}
int N, M, Q;
bool arr[1100][1100];
int dp[1100][1100][2], ans;
void upd(int x, int y, int d) {
  if (x < 1 || y < 1) return;
  if (arr[x][y]) return;
  if (d == 0) {
    ans -= dp[x][y][d];
    dp[x][y][d] = dp[x][y + 1][1 - d];
    if (y + 1 <= M && !arr[x][y + 1]) dp[x][y][d]++;
    ans += dp[x][y][d];
    upd(x - 1, y, 1 - d);
  } else {
    ans -= dp[x][y][d];
    dp[x][y][d] = dp[x + 1][y][1 - d];
    if (x + 1 <= N && !arr[x + 1][y]) dp[x][y][d]++;
    ans += dp[x][y][d];
    upd(x, y - 1, 1 - d);
  }
}
int main() {
  fastIO();
  cin >> N >> M >> Q;
  for (int i = N; i >= 1; i--) {
    for (int j = M; j >= 1; j--) {
      if (j + 1 <= M) dp[i][j][0] = dp[i][j + 1][1] + 1;
      if (i + 1 <= N) dp[i][j][1] = dp[i + 1][j][0] + 1;
    }
  }
  ans = N * M;
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= M; j++) {
      ans += dp[i][j][0] + dp[i][j][1];
    }
  }
  for (int i = 1; i <= Q; i++) {
    int a, b;
    cin >> a >> b;
    if (arr[a][b] == false) {
      ans -= dp[a][b][0] + dp[a][b][1];
      dp[a][b][0] = 0;
      dp[a][b][1] = 0;
      ans--;
      arr[a][b] = true;
      upd(a, b - 1, 0);
      upd(a - 1, b, 1);
    } else {
      ans++;
      arr[a][b] = false;
      upd(a, b, 0);
      upd(a, b, 1);
    }
    cout << ans << "\n";
  }
}

```