#include <bits/stdc++.h>
using namespace std;
int ttt = 1;
long long n, m, k;
template <class myType>
void print_arr(myType &arr, long long L, string sep) {
  for (int(i) = (0); (i) < (L); ++(i)) {
    cout << arr[i];
    if (i < L - 1) cout << sep;
  }
  cout << "\n";
  return;
}
void solve() {
  cin >> n >> m >> k;
  long long ans = 0;
  for (int(i) = (0); (i) < (n); ++(i))
    for (int(j) = (0); (j) < (m); ++(j)) {
      ans++;
      long long d = n - i - 1, r = m - j - 1;
      ans += 2 * min(d, r) + (r > d ? 1 : 0);
      ans += 2 * min(d, r) + (d > r ? 1 : 0);
    }
  vector<vector<int> > G(n, vector<int>(m, 0));
  while (k--) {
    int i, j;
    cin >> i >> j;
    i--;
    j--;
    if (G[i][j] == 1)
      ans++;
    else
      ans--;
    int r_then_d = 1;
    int x = i, y = j;
    while (1) {
      y++;
      if (y >= m || G[x][y] == 1) break;
      r_then_d++;
      x++;
      if (x >= n || G[x][y] == 1) break;
      r_then_d++;
    }
    int d_then_r = 1;
    x = i;
    y = j;
    while (1) {
      x++;
      if (x >= n || G[x][y] == 1) break;
      d_then_r++;
      y++;
      if (y >= m || G[x][y] == 1) break;
      d_then_r++;
    }
    int u_then_l = 1;
    x = i;
    y = j;
    while (1) {
      x--;
      if (x < 0 || G[x][y] == 1) break;
      u_then_l++;
      y--;
      if (y < 0 || G[x][y] == 1) break;
      u_then_l++;
    }
    int l_then_u = 1;
    x = i;
    y = j;
    while (1) {
      y--;
      if (y < 0 || G[x][y] == 1) break;
      l_then_u++;
      x--;
      if (x < 0 || G[x][y] == 1) break;
      l_then_u++;
    }
    if (G[i][j] == 1) {
      ans += u_then_l * r_then_d - 1 + l_then_u * d_then_r - 1;
    } else {
      ans -= u_then_l * r_then_d - 1 + l_then_u * d_then_r - 1;
    }
    cout << ans << "\n";
    G[i][j] ^= 1;
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  while (ttt--) solve();
  return 0;
}

### 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 ttt = 1;
long long n, m, k;
template <class myType>
void print_arr(myType &arr, long long L, string sep) {
  for (int(i) = (0); (i) < (L); ++(i)) {
    cout << arr[i];
    if (i < L - 1) cout << sep;
  }
  cout << "\n";
  return;
}
void solve() {
  cin >> n >> m >> k;
  long long ans = 0;
  for (int(i) = (0); (i) < (n); ++(i))
    for (int(j) = (0); (j) < (m); ++(j)) {
      ans++;
      long long d = n - i - 1, r = m - j - 1;
      ans += 2 * min(d, r) + (r > d ? 1 : 0);
      ans += 2 * min(d, r) + (d > r ? 1 : 0);
    }
  vector<vector<int> > G(n, vector<int>(m, 0));
  while (k--) {
    int i, j;
    cin >> i >> j;
    i--;
    j--;
    if (G[i][j] == 1)
      ans++;
    else
      ans--;
    int r_then_d = 1;
    int x = i, y = j;
    while (1) {
      y++;
      if (y >= m || G[x][y] == 1) break;
      r_then_d++;
      x++;
      if (x >= n || G[x][y] == 1) break;
      r_then_d++;
    }
    int d_then_r = 1;
    x = i;
    y = j;
    while (1) {
      x++;
      if (x >= n || G[x][y] == 1) break;
      d_then_r++;
      y++;
      if (y >= m || G[x][y] == 1) break;
      d_then_r++;
    }
    int u_then_l = 1;
    x = i;
    y = j;
    while (1) {
      x--;
      if (x < 0 || G[x][y] == 1) break;
      u_then_l++;
      y--;
      if (y < 0 || G[x][y] == 1) break;
      u_then_l++;
    }
    int l_then_u = 1;
    x = i;
    y = j;
    while (1) {
      y--;
      if (y < 0 || G[x][y] == 1) break;
      l_then_u++;
      x--;
      if (x < 0 || G[x][y] == 1) break;
      l_then_u++;
    }
    if (G[i][j] == 1) {
      ans += u_then_l * r_then_d - 1 + l_then_u * d_then_r - 1;
    } else {
      ans -= u_then_l * r_then_d - 1 + l_then_u * d_then_r - 1;
    }
    cout << ans << "\n";
    G[i][j] ^= 1;
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  while (ttt--) solve();
  return 0;
}

```