#include <bits/stdc++.h>
using namespace std;
long long n, m, q, dp[1005][1005][2], flag[1005][1005];
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      dp[i][j][0] = dp[i][j][1] = 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (i > 1) dp[i][j][0] = 1 + dp[i - 1][j][1];
      if (j > 1) dp[i][j][1] = 1 + dp[i][j - 1][0];
      flag[i][j] = 1;
    }
  }
  long long res = 0, tot = n * m;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      res += dp[i][j][0] + dp[i][j][1];
    }
  }
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    flag[x][y] ^= 1;
    if (flag[x][y] == 0) {
      tot--;
      res -= dp[x][y][0] + dp[x][y][1];
      dp[x][y][0] = dp[x][y][1] = 0;
      res += (dp[x][y][0] + dp[x][y][1]);
    } else {
      tot++;
      res -= dp[x][y][0] + dp[x][y][1];
      if (x > 1)
        dp[x][y][0] = dp[x - 1][y][1] + 1;
      else
        dp[x][y][0] = 1;
      if (y > 1)
        dp[x][y][1] = dp[x][y - 1][0] + 1;
      else
        dp[x][y][1] = 1;
      res += dp[x][y][1] + dp[x][y][0];
    }
    while (x <= n && y <= m) {
      if (x < n) {
        res -= dp[x + 1][y][0];
        dp[x + 1][y][0] = flag[x + 1][y] * (1 + dp[x][y][1]);
        res += dp[x + 1][y][0];
      }
      if (y < m) {
        res -= dp[x][y + 1][1];
        dp[x][y + 1][1] = flag[x][y + 1] * (1 + dp[x][y][0]);
        res += dp[x][y + 1][1];
      }
      if (x < n && y < m) {
        res -= dp[x + 1][y + 1][1] + dp[x + 1][y + 1][0];
        dp[x + 1][y + 1][0] = flag[x + 1][y + 1] * (1 + dp[x][y + 1][1]);
        dp[x + 1][y + 1][1] = flag[x + 1][y + 1] * (1 + dp[x + 1][y][0]);
        res += dp[x + 1][y + 1][1] + dp[x + 1][y + 1][0];
      }
      x++, y++;
    }
    printf("%lld\n", res - tot);
  }
}

### 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;
long long n, m, q, dp[1005][1005][2], flag[1005][1005];
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      dp[i][j][0] = dp[i][j][1] = 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (i > 1) dp[i][j][0] = 1 + dp[i - 1][j][1];
      if (j > 1) dp[i][j][1] = 1 + dp[i][j - 1][0];
      flag[i][j] = 1;
    }
  }
  long long res = 0, tot = n * m;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      res += dp[i][j][0] + dp[i][j][1];
    }
  }
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    flag[x][y] ^= 1;
    if (flag[x][y] == 0) {
      tot--;
      res -= dp[x][y][0] + dp[x][y][1];
      dp[x][y][0] = dp[x][y][1] = 0;
      res += (dp[x][y][0] + dp[x][y][1]);
    } else {
      tot++;
      res -= dp[x][y][0] + dp[x][y][1];
      if (x > 1)
        dp[x][y][0] = dp[x - 1][y][1] + 1;
      else
        dp[x][y][0] = 1;
      if (y > 1)
        dp[x][y][1] = dp[x][y - 1][0] + 1;
      else
        dp[x][y][1] = 1;
      res += dp[x][y][1] + dp[x][y][0];
    }
    while (x <= n && y <= m) {
      if (x < n) {
        res -= dp[x + 1][y][0];
        dp[x + 1][y][0] = flag[x + 1][y] * (1 + dp[x][y][1]);
        res += dp[x + 1][y][0];
      }
      if (y < m) {
        res -= dp[x][y + 1][1];
        dp[x][y + 1][1] = flag[x][y + 1] * (1 + dp[x][y][0]);
        res += dp[x][y + 1][1];
      }
      if (x < n && y < m) {
        res -= dp[x + 1][y + 1][1] + dp[x + 1][y + 1][0];
        dp[x + 1][y + 1][0] = flag[x + 1][y + 1] * (1 + dp[x][y + 1][1]);
        dp[x + 1][y + 1][1] = flag[x + 1][y + 1] * (1 + dp[x + 1][y][0]);
        res += dp[x + 1][y + 1][1] + dp[x + 1][y + 1][0];
      }
      x++, y++;
    }
    printf("%lld\n", res - tot);
  }
}

```