#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n, m;
  cin >> n >> m;
  vector<vector<vector<int>>> dp(n + 1,
                                 vector<vector<int>>(m + 1, vector<int>(2, 0)));
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      dp[i][j][1] = dp[i][j - 1][0] + 1;
      dp[i][j][0] = dp[i - 1][j][1] + 1;
    }
  int ans = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      ans += dp[i][j][0] + dp[i][j][1];
    }
  ans -= n * m;
  int q;
  cin >> q;
  vector<vector<int>> a(n, vector<int>(m, 1));
  auto calc_len = [&](int x, int y, int dx, int dy) {
    int res = 0;
    while (1) {
      x += dx;
      y += dy;
      if (x < 0 || y < 0 || x >= n || y >= m || a[x][y] == 0) {
        break;
      }
      res++;
      swap(dx, dy);
    }
    return res;
  };
  while (q--) {
    int i, j;
    cin >> i >> j;
    i--, j--;
    {
      int x = calc_len(i, j, -1, 0);
      int y = calc_len(i, j, 0, 1);
      ans += (a[i][j] == 1 ? -1 : 1) * ((x + 1) * (y + 1));
    }
    {
      int x = calc_len(i, j, 0, -1);
      int y = calc_len(i, j, 1, 0);
      ans += (a[i][j] == 1 ? -1 : 1) * ((x + 1) * (y + 1));
    }
    ans += (a[i][j] == 1 ? 1 : -1);
    a[i][j] ^= 1;
    cout << ans << endl;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int T = 1;
  while (T--) {
    solve();
  }
}

### Prompt

Please formulate 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;
void solve() {
  int n, m;
  cin >> n >> m;
  vector<vector<vector<int>>> dp(n + 1,
                                 vector<vector<int>>(m + 1, vector<int>(2, 0)));
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      dp[i][j][1] = dp[i][j - 1][0] + 1;
      dp[i][j][0] = dp[i - 1][j][1] + 1;
    }
  int ans = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      ans += dp[i][j][0] + dp[i][j][1];
    }
  ans -= n * m;
  int q;
  cin >> q;
  vector<vector<int>> a(n, vector<int>(m, 1));
  auto calc_len = [&](int x, int y, int dx, int dy) {
    int res = 0;
    while (1) {
      x += dx;
      y += dy;
      if (x < 0 || y < 0 || x >= n || y >= m || a[x][y] == 0) {
        break;
      }
      res++;
      swap(dx, dy);
    }
    return res;
  };
  while (q--) {
    int i, j;
    cin >> i >> j;
    i--, j--;
    {
      int x = calc_len(i, j, -1, 0);
      int y = calc_len(i, j, 0, 1);
      ans += (a[i][j] == 1 ? -1 : 1) * ((x + 1) * (y + 1));
    }
    {
      int x = calc_len(i, j, 0, -1);
      int y = calc_len(i, j, 1, 0);
      ans += (a[i][j] == 1 ? -1 : 1) * ((x + 1) * (y + 1));
    }
    ans += (a[i][j] == 1 ? 1 : -1);
    a[i][j] ^= 1;
    cout << ans << endl;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int T = 1;
  while (T--) {
    solve();
  }
}

```