#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long int n, m, q;
  cin >> n >> m >> q;
  vector<vector<bool>> grid(n + 2, vector<bool>(m + 2, true));
  vector<vector<long long int>> one(n + 2, vector<long long int>(m + 2, 0));
  vector<vector<long long int>> two(n + 2, vector<long long int>(m + 2, 0));
  long long int total = 0;
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = 1; j <= m; j++) {
      one[i][j] = two[i][j] = 1;
      one[i][j] += two[i][j - 1];
      two[i][j] += one[i - 1][j];
      total += one[i][j] + two[i][j] - 1;
    }
  }
  for (long long int i = 1; i <= q; i++) {
    long long int x, y;
    cin >> x >> y;
    if (grid[x][y]) {
      grid[x][y] = !grid[x][y];
      total -= one[x][y];
      total -= two[x][y];
      total++;
      one[x][y] = 0;
      two[x][y] = 0;
      for (long long int step = 1; step <= max(n, m); step++) {
        if ((x + step - 1 <= n) and (y + step <= m) &&
            grid[x + step - 1][y + step]) {
          long long int temp = two[x + step - 1][y + step - 1] + 1;
          total += (temp - one[x + step - 1][y + step]);
          one[x + step - 1][y + step] = temp;
        }
        if ((x + step <= n) and (y + step - 1 <= m) &&
            grid[x + step][y + step - 1]) {
          long long int temp = one[x + step - 1][y + step - 1] + 1;
          total += (temp - two[x + step][y + step - 1]);
          two[x + step][y + step - 1] = temp;
        }
        if (x + step > n || y + step > m) break;
        if (!grid[x + step][y + step]) break;
        long long int temp = one[x + step - 1][y + step] + 1;
        total += (temp - two[x + step][y + step]);
        two[x + step][y + step] = temp;
        temp = two[x + step][y + step - 1] + 1;
        total += (temp - one[x + step][y + step]);
        one[x + step][y + step] = temp;
      }
    } else {
      grid[x][y] = !grid[x][y];
      one[x][y] = 1 + two[x][y - 1];
      two[x][y] = 1 + one[x - 1][y];
      total += one[x][y] + two[x][y] - 1;
      for (long long int step = 1; step <= max(n, m); step++) {
        if ((x + step - 1 <= n) and (y + step <= m) &&
            grid[x + step - 1][y + step]) {
          long long int temp = two[x + step - 1][y + step - 1] + 1;
          total += (temp - one[x + step - 1][y + step]);
          one[x + step - 1][y + step] = temp;
        }
        if ((x + step <= n) and (y + step - 1 <= m) &&
            grid[x + step][y + step - 1]) {
          long long int temp = one[x + step - 1][y + step - 1] + 1;
          total += (temp - two[x + step][y + step - 1]);
          two[x + step][y + step - 1] = temp;
        }
        if (x + step > n || y + step > m) break;
        if (!grid[x + step][y + step]) break;
        long long int temp = one[x + step - 1][y + step] + 1;
        total += (temp - two[x + step][y + step]);
        two[x + step][y + step] = temp;
        temp = two[x + step][y + step - 1] + 1;
        total += (temp - one[x + step][y + step]);
        one[x + step][y + step] = temp;
      }
    }
    cout << total << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long tc = 1;
  while (tc--) {
    solve();
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 solve() {
  long long int n, m, q;
  cin >> n >> m >> q;
  vector<vector<bool>> grid(n + 2, vector<bool>(m + 2, true));
  vector<vector<long long int>> one(n + 2, vector<long long int>(m + 2, 0));
  vector<vector<long long int>> two(n + 2, vector<long long int>(m + 2, 0));
  long long int total = 0;
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = 1; j <= m; j++) {
      one[i][j] = two[i][j] = 1;
      one[i][j] += two[i][j - 1];
      two[i][j] += one[i - 1][j];
      total += one[i][j] + two[i][j] - 1;
    }
  }
  for (long long int i = 1; i <= q; i++) {
    long long int x, y;
    cin >> x >> y;
    if (grid[x][y]) {
      grid[x][y] = !grid[x][y];
      total -= one[x][y];
      total -= two[x][y];
      total++;
      one[x][y] = 0;
      two[x][y] = 0;
      for (long long int step = 1; step <= max(n, m); step++) {
        if ((x + step - 1 <= n) and (y + step <= m) &&
            grid[x + step - 1][y + step]) {
          long long int temp = two[x + step - 1][y + step - 1] + 1;
          total += (temp - one[x + step - 1][y + step]);
          one[x + step - 1][y + step] = temp;
        }
        if ((x + step <= n) and (y + step - 1 <= m) &&
            grid[x + step][y + step - 1]) {
          long long int temp = one[x + step - 1][y + step - 1] + 1;
          total += (temp - two[x + step][y + step - 1]);
          two[x + step][y + step - 1] = temp;
        }
        if (x + step > n || y + step > m) break;
        if (!grid[x + step][y + step]) break;
        long long int temp = one[x + step - 1][y + step] + 1;
        total += (temp - two[x + step][y + step]);
        two[x + step][y + step] = temp;
        temp = two[x + step][y + step - 1] + 1;
        total += (temp - one[x + step][y + step]);
        one[x + step][y + step] = temp;
      }
    } else {
      grid[x][y] = !grid[x][y];
      one[x][y] = 1 + two[x][y - 1];
      two[x][y] = 1 + one[x - 1][y];
      total += one[x][y] + two[x][y] - 1;
      for (long long int step = 1; step <= max(n, m); step++) {
        if ((x + step - 1 <= n) and (y + step <= m) &&
            grid[x + step - 1][y + step]) {
          long long int temp = two[x + step - 1][y + step - 1] + 1;
          total += (temp - one[x + step - 1][y + step]);
          one[x + step - 1][y + step] = temp;
        }
        if ((x + step <= n) and (y + step - 1 <= m) &&
            grid[x + step][y + step - 1]) {
          long long int temp = one[x + step - 1][y + step - 1] + 1;
          total += (temp - two[x + step][y + step - 1]);
          two[x + step][y + step - 1] = temp;
        }
        if (x + step > n || y + step > m) break;
        if (!grid[x + step][y + step]) break;
        long long int temp = one[x + step - 1][y + step] + 1;
        total += (temp - two[x + step][y + step]);
        two[x + step][y + step] = temp;
        temp = two[x + step][y + step - 1] + 1;
        total += (temp - one[x + step][y + step]);
        one[x + step][y + step] = temp;
      }
    }
    cout << total << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long tc = 1;
  while (tc--) {
    solve();
  }
  return 0;
}

```