#include <bits/stdc++.h>
using namespace std;
int n, m, q;
int total;
int n_to_end(int row_start, int col_start, int dir, int dir_next,
             vector<vector<int> >& field) {
  if (row_start < 0 || row_start >= n) return 0;
  if (col_start < 0 || col_start >= m) return 0;
  int next_row = row_start;
  int next_col = col_start;
  if (dir == 1) {
    next_row -= 1;
  }
  if (dir == 2) {
    next_col += 1;
  }
  if (dir == 3) {
    next_row += 1;
  }
  if (dir == 4) {
    next_col -= 1;
  }
  if (next_row < 0 || next_row >= n) return 1;
  if (next_col < 0 || next_col >= m) return 1;
  if (field[next_row][next_col] == 1) return 1;
  return 1 + n_to_end(next_row, next_col, dir_next, dir, field);
}
int n_length_2_up(int length) { return ((length - 1) * length) / 2; }
int main() {
  cin >> n >> m >> q;
  vector<vector<int> > field(n, vector<int>(m, 0));
  int total = 0;
  for (int i = 0; i < n - 1; i++) {
    int length = n_to_end(i, 0, 3, 2, field);
    total += n_length_2_up(length);
  }
  for (int k = 0; k < m - 1; k++) {
    int length = n_to_end(0, k, 2, 3, field);
    total += n_length_2_up(length);
  }
  total += n * m;
  for (int i = 0; i < q; i++) {
    int row, col;
    cin >> row >> col;
    row--;
    col--;
    int length1 = n_to_end(row, col, 1, 4, field);
    int length2 = n_to_end(row, col, 2, 3, field);
    int length3 = n_to_end(row, col, 3, 2, field);
    int length4 = n_to_end(row, col, 4, 1, field);
    int sum = ((length1 * length2) - 1) + ((length3 * length4) - 1) + 1;
    if (field[row][col]) {
      total += sum;
    } else {
      total -= sum;
    }
    cout << total << '\n';
    field[row][col] ^= 1;
  }
}

### Prompt

Your task is to create 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;
int n, m, q;
int total;
int n_to_end(int row_start, int col_start, int dir, int dir_next,
             vector<vector<int> >& field) {
  if (row_start < 0 || row_start >= n) return 0;
  if (col_start < 0 || col_start >= m) return 0;
  int next_row = row_start;
  int next_col = col_start;
  if (dir == 1) {
    next_row -= 1;
  }
  if (dir == 2) {
    next_col += 1;
  }
  if (dir == 3) {
    next_row += 1;
  }
  if (dir == 4) {
    next_col -= 1;
  }
  if (next_row < 0 || next_row >= n) return 1;
  if (next_col < 0 || next_col >= m) return 1;
  if (field[next_row][next_col] == 1) return 1;
  return 1 + n_to_end(next_row, next_col, dir_next, dir, field);
}
int n_length_2_up(int length) { return ((length - 1) * length) / 2; }
int main() {
  cin >> n >> m >> q;
  vector<vector<int> > field(n, vector<int>(m, 0));
  int total = 0;
  for (int i = 0; i < n - 1; i++) {
    int length = n_to_end(i, 0, 3, 2, field);
    total += n_length_2_up(length);
  }
  for (int k = 0; k < m - 1; k++) {
    int length = n_to_end(0, k, 2, 3, field);
    total += n_length_2_up(length);
  }
  total += n * m;
  for (int i = 0; i < q; i++) {
    int row, col;
    cin >> row >> col;
    row--;
    col--;
    int length1 = n_to_end(row, col, 1, 4, field);
    int length2 = n_to_end(row, col, 2, 3, field);
    int length3 = n_to_end(row, col, 3, 2, field);
    int length4 = n_to_end(row, col, 4, 1, field);
    int sum = ((length1 * length2) - 1) + ((length3 * length4) - 1) + 1;
    if (field[row][col]) {
      total += sum;
    } else {
      total -= sum;
    }
    cout << total << '\n';
    field[row][col] ^= 1;
  }
}

```