#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m, q;
  cin >> n >> m >> q;
  vector<int> x(q), y(q);
  for (int i = (0); i < (q); i++) {
    cin >> x[i] >> y[i];
    x[i]--;
    y[i]--;
  }
  vector<vector<long long>> u(n, vector<long long>(m, 1)),
      l(n, vector<long long>(m, 1));
  long long ans = 0, t = 0;
  for (int i = (0); i < (n); i++) {
    for (int j = (0); j < (m); j++) {
      if (i > 0) u[i][j] = l[i - 1][j] + 1;
      if (j > 0) l[i][j] = u[i][j - 1] + 1;
      ans += u[i][j] + l[i][j];
      t++;
    }
  }
  for (int i = (0); i < (q); i++) {
    if (u[x[i]][y[i]] > 0) {
      ans -= u[x[i]][y[i]] + l[x[i]][y[i]];
      t--;
      u[x[i]][y[i]] = 0;
      l[x[i]][y[i]] = 0;
    } else {
      t++;
      u[x[i]][y[i]] = 1;
      l[x[i]][y[i]] = 1;
      if (x[i] > 0) u[x[i]][y[i]] = l[x[i] - 1][y[i]] + 1;
      if (y[i] > 0) l[x[i]][y[i]] = u[x[i]][y[i] - 1] + 1;
      ans += u[x[i]][y[i]] + l[x[i]][y[i]];
    }
    for (int j = (x[i]); j < (n); j++) {
      for (int k = (y[i] + j - x[i] - 1); k < (y[i] + j - x[i] + 2); k++) {
        if (j == x[i]) {
          if (k < y[i] + j - x[i] + 1) continue;
        }
        if (k >= m) continue;
        if (u[j][k] == 0) continue;
        ans -= u[j][k] + l[j][k];
        u[j][k] = 1;
        l[j][k] = 1;
        if (j > 0) u[j][k] = l[j - 1][k] + 1;
        if (k > 0) l[j][k] = u[j][k - 1] + 1;
        ans += u[j][k] + l[j][k];
      }
    }
    cout << ans - t << endl;
  }
}

### Prompt

Generate 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 main() {
  int n, m, q;
  cin >> n >> m >> q;
  vector<int> x(q), y(q);
  for (int i = (0); i < (q); i++) {
    cin >> x[i] >> y[i];
    x[i]--;
    y[i]--;
  }
  vector<vector<long long>> u(n, vector<long long>(m, 1)),
      l(n, vector<long long>(m, 1));
  long long ans = 0, t = 0;
  for (int i = (0); i < (n); i++) {
    for (int j = (0); j < (m); j++) {
      if (i > 0) u[i][j] = l[i - 1][j] + 1;
      if (j > 0) l[i][j] = u[i][j - 1] + 1;
      ans += u[i][j] + l[i][j];
      t++;
    }
  }
  for (int i = (0); i < (q); i++) {
    if (u[x[i]][y[i]] > 0) {
      ans -= u[x[i]][y[i]] + l[x[i]][y[i]];
      t--;
      u[x[i]][y[i]] = 0;
      l[x[i]][y[i]] = 0;
    } else {
      t++;
      u[x[i]][y[i]] = 1;
      l[x[i]][y[i]] = 1;
      if (x[i] > 0) u[x[i]][y[i]] = l[x[i] - 1][y[i]] + 1;
      if (y[i] > 0) l[x[i]][y[i]] = u[x[i]][y[i] - 1] + 1;
      ans += u[x[i]][y[i]] + l[x[i]][y[i]];
    }
    for (int j = (x[i]); j < (n); j++) {
      for (int k = (y[i] + j - x[i] - 1); k < (y[i] + j - x[i] + 2); k++) {
        if (j == x[i]) {
          if (k < y[i] + j - x[i] + 1) continue;
        }
        if (k >= m) continue;
        if (u[j][k] == 0) continue;
        ans -= u[j][k] + l[j][k];
        u[j][k] = 1;
        l[j][k] = 1;
        if (j > 0) u[j][k] = l[j - 1][k] + 1;
        if (k > 0) l[j][k] = u[j][k - 1] + 1;
        ans += u[j][k] + l[j][k];
      }
    }
    cout << ans - t << endl;
  }
}

```