#include <bits/stdc++.h>
using namespace std;
long long int arr[1001][1001];
long long int dp[1001][1001];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n, m, q;
  cin >> n >> m >> q;
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= m; j++) arr[i][j] = 1, dp[i][j] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      long long int d = n - i, r = m - j;
      arr[i][j] += (4 * min(d, r));
      if (d != r) arr[i][j]++;
    }
  }
  long long int sum = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      sum += arr[i][j];
    }
  }
  while (q--) {
    long long int x, y;
    cin >> x >> y;
    long long int r = x, c = y;
    if (dp[r][c] == 0) {
      long long int temp = 0;
      long long int cnt = 0;
      while (r <= n && c <= m && dp[r][c] == 0) {
        temp++;
        if (cnt % 2 == 0)
          r++;
        else
          c++;
        cnt++;
      }
      r = x, c = y;
      cnt = 0;
      while (r >= 1 && c >= 1 && dp[r][c] == 0) {
        arr[r][c] -= temp;
        sum -= temp;
        if (cnt % 2 == 0)
          c--;
        else
          r--;
        cnt++;
      }
      temp = 0;
      cnt = 0;
      r = x, c = y;
      while (r <= n && c <= m && dp[r][c] == 0) {
        temp++;
        if (cnt % 2 == 0)
          c++;
        else
          r++;
        cnt++;
      }
      r = x, c = y;
      cnt = 0;
      while (r >= 1 && c >= 1 && dp[r][c] == 0) {
        arr[r][c] -= temp;
        sum -= temp;
        if (cnt % 2 == 0)
          r--;
        else
          c--;
        cnt++;
      }
      arr[x][y]++;
      sum += 1;
      dp[x][y] = 1;
      cout << sum << endl;
    } else {
      dp[x][y] = 0;
      long long int temp = 0;
      long long int cnt = 0;
      while (r <= n && c <= m && dp[r][c] == 0) {
        temp++;
        if (cnt % 2 == 0)
          r++;
        else
          c++;
        cnt++;
      }
      r = x, c = y;
      cnt = 0;
      while (r >= 1 && c >= 1 && dp[r][c] == 0) {
        arr[r][c] += temp;
        sum += temp;
        if (cnt % 2 == 0)
          c--;
        else
          r--;
        cnt++;
      }
      temp = 0;
      cnt = 0;
      r = x, c = y;
      while (r <= n && c <= m && dp[r][c] == 0) {
        temp++;
        if (cnt % 2 == 0)
          c++;
        else
          r++;
        cnt++;
      }
      r = x, c = y;
      cnt = 0;
      while (r >= 1 && c >= 1 && dp[r][c] == 0) {
        arr[r][c] += temp;
        sum += temp;
        if (cnt % 2 == 0)
          r--;
        else
          c--;
        cnt++;
      }
      arr[x][y]--;
      sum -= 1;
      cout << sum << endl;
    }
  }
}

### Prompt

Your challenge is to write 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;
long long int arr[1001][1001];
long long int dp[1001][1001];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n, m, q;
  cin >> n >> m >> q;
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= m; j++) arr[i][j] = 1, dp[i][j] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      long long int d = n - i, r = m - j;
      arr[i][j] += (4 * min(d, r));
      if (d != r) arr[i][j]++;
    }
  }
  long long int sum = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      sum += arr[i][j];
    }
  }
  while (q--) {
    long long int x, y;
    cin >> x >> y;
    long long int r = x, c = y;
    if (dp[r][c] == 0) {
      long long int temp = 0;
      long long int cnt = 0;
      while (r <= n && c <= m && dp[r][c] == 0) {
        temp++;
        if (cnt % 2 == 0)
          r++;
        else
          c++;
        cnt++;
      }
      r = x, c = y;
      cnt = 0;
      while (r >= 1 && c >= 1 && dp[r][c] == 0) {
        arr[r][c] -= temp;
        sum -= temp;
        if (cnt % 2 == 0)
          c--;
        else
          r--;
        cnt++;
      }
      temp = 0;
      cnt = 0;
      r = x, c = y;
      while (r <= n && c <= m && dp[r][c] == 0) {
        temp++;
        if (cnt % 2 == 0)
          c++;
        else
          r++;
        cnt++;
      }
      r = x, c = y;
      cnt = 0;
      while (r >= 1 && c >= 1 && dp[r][c] == 0) {
        arr[r][c] -= temp;
        sum -= temp;
        if (cnt % 2 == 0)
          r--;
        else
          c--;
        cnt++;
      }
      arr[x][y]++;
      sum += 1;
      dp[x][y] = 1;
      cout << sum << endl;
    } else {
      dp[x][y] = 0;
      long long int temp = 0;
      long long int cnt = 0;
      while (r <= n && c <= m && dp[r][c] == 0) {
        temp++;
        if (cnt % 2 == 0)
          r++;
        else
          c++;
        cnt++;
      }
      r = x, c = y;
      cnt = 0;
      while (r >= 1 && c >= 1 && dp[r][c] == 0) {
        arr[r][c] += temp;
        sum += temp;
        if (cnt % 2 == 0)
          c--;
        else
          r--;
        cnt++;
      }
      temp = 0;
      cnt = 0;
      r = x, c = y;
      while (r <= n && c <= m && dp[r][c] == 0) {
        temp++;
        if (cnt % 2 == 0)
          c++;
        else
          r++;
        cnt++;
      }
      r = x, c = y;
      cnt = 0;
      while (r >= 1 && c >= 1 && dp[r][c] == 0) {
        arr[r][c] += temp;
        sum += temp;
        if (cnt % 2 == 0)
          r--;
        else
          c--;
        cnt++;
      }
      arr[x][y]--;
      sum -= 1;
      cout << sum << endl;
    }
  }
}

```