#include <bits/stdc++.h>
using namespace std;
auto fio = (ios::sync_with_stdio(0), cin.tie(0), cout.tie(0));
long long G[1010][1010];
int32_t main() {
  long long n, m;
  cin >> n >> m;
  auto isValid = [&](long long i, long long j) {
    return i >= 0 && i < n && j >= 0 && j < m && !G[i][j];
  };
  long long ans = 0;
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j < m; j++) ans += min(i, j) * 4 + 1 + (i != j);
  long long q;
  cin >> q;
  while (q--) {
    long long x, y;
    cin >> x >> y;
    x--, y--;
    long long a = 0, b = 0;
    long long c = 0, d = 0;
    long long old = G[x][y];
    G[x][y] = 0;
    for (long long i = x, j = y; isValid(i, j); a++)
      if (a & 1)
        i--;
      else
        j--;
    for (long long i = x, j = y; isValid(i, j); b++)
      if (b & 1)
        j++;
      else
        i++;
    for (long long i = x, j = y; isValid(i, j); c++)
      if (c & 1)
        j--;
      else
        i--;
    for (long long i = x, j = y; isValid(i, j); d++)
      if (d & 1)
        i++;
      else
        j++;
    long long w = a * b + c * d - 1;
    if (old)
      ans += w;
    else
      ans -= w;
    G[x][y] = !old;
    cout << ans << endl;
  }
}

### Prompt

Create a solution in CPP for 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;
auto fio = (ios::sync_with_stdio(0), cin.tie(0), cout.tie(0));
long long G[1010][1010];
int32_t main() {
  long long n, m;
  cin >> n >> m;
  auto isValid = [&](long long i, long long j) {
    return i >= 0 && i < n && j >= 0 && j < m && !G[i][j];
  };
  long long ans = 0;
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j < m; j++) ans += min(i, j) * 4 + 1 + (i != j);
  long long q;
  cin >> q;
  while (q--) {
    long long x, y;
    cin >> x >> y;
    x--, y--;
    long long a = 0, b = 0;
    long long c = 0, d = 0;
    long long old = G[x][y];
    G[x][y] = 0;
    for (long long i = x, j = y; isValid(i, j); a++)
      if (a & 1)
        i--;
      else
        j--;
    for (long long i = x, j = y; isValid(i, j); b++)
      if (b & 1)
        j++;
      else
        i++;
    for (long long i = x, j = y; isValid(i, j); c++)
      if (c & 1)
        j--;
      else
        i--;
    for (long long i = x, j = y; isValid(i, j); d++)
      if (d & 1)
        i++;
      else
        j++;
    long long w = a * b + c * d - 1;
    if (old)
      ans += w;
    else
      ans -= w;
    G[x][y] = !old;
    cout << ans << endl;
  }
}

```