#include <bits/stdc++.h>
using namespace std;
void solve() {
  using ll = long long;
  ll n, m, q;
  cin >> n >> m >> q;
  vector<vector<bool>> matrix(n, vector<bool>(m, true));
  ll count = -m * n;
  for (ll i = 0; i < n; i++) {
    if (m > i) count += 2 * (n - i) * (m - i);
    if (m > i + 1) count += (n - i) * (m - i - 1);
    if (i > 0 && m > i - 1) count += (n - i) * (m - i + 1);
  }
  while (q--) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    ll changeCount = 1;
    int cx, cy;
    ll leftCount = 0;
    bool turn = 1;
    cx = x - 1;
    cy = y;
    for (; cx >= 0 && cy >= 0; turn ? cx-- : cy--) {
      if (!matrix[cx][cy]) break;
      leftCount++;
      turn = !turn;
    }
    ll bottomCount = 0;
    turn = 1;
    cx = x;
    cy = y + 1;
    for (; cx < n && cy < m; turn ? cy++ : cx++) {
      if (!matrix[cx][cy]) break;
      bottomCount++;
      turn = !turn;
    }
    changeCount += (leftCount + 1) * (bottomCount + 1) - 1;
    ll topCount = 0;
    turn = 1;
    cx = x;
    cy = y - 1;
    for (; cx >= 0 && cy >= 0; turn ? cy-- : cx--) {
      if (!matrix[cx][cy]) break;
      topCount++;
      turn = !turn;
    }
    ll rightCount = 0;
    turn = 1;
    cx = x + 1;
    cy = y;
    for (; cx < n && cy < m; turn ? cx++ : cy++) {
      if (!matrix[cx][cy]) break;
      rightCount++;
      turn = !turn;
    }
    changeCount += (topCount + 1) * (rightCount + 1) - 1;
    if (matrix[x][y]) {
      count -= changeCount;
    } else {
      count += changeCount;
    }
    matrix[x][y] = !matrix[x][y];
    cout << count << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  solve();
}

### 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;
void solve() {
  using ll = long long;
  ll n, m, q;
  cin >> n >> m >> q;
  vector<vector<bool>> matrix(n, vector<bool>(m, true));
  ll count = -m * n;
  for (ll i = 0; i < n; i++) {
    if (m > i) count += 2 * (n - i) * (m - i);
    if (m > i + 1) count += (n - i) * (m - i - 1);
    if (i > 0 && m > i - 1) count += (n - i) * (m - i + 1);
  }
  while (q--) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    ll changeCount = 1;
    int cx, cy;
    ll leftCount = 0;
    bool turn = 1;
    cx = x - 1;
    cy = y;
    for (; cx >= 0 && cy >= 0; turn ? cx-- : cy--) {
      if (!matrix[cx][cy]) break;
      leftCount++;
      turn = !turn;
    }
    ll bottomCount = 0;
    turn = 1;
    cx = x;
    cy = y + 1;
    for (; cx < n && cy < m; turn ? cy++ : cx++) {
      if (!matrix[cx][cy]) break;
      bottomCount++;
      turn = !turn;
    }
    changeCount += (leftCount + 1) * (bottomCount + 1) - 1;
    ll topCount = 0;
    turn = 1;
    cx = x;
    cy = y - 1;
    for (; cx >= 0 && cy >= 0; turn ? cy-- : cx--) {
      if (!matrix[cx][cy]) break;
      topCount++;
      turn = !turn;
    }
    ll rightCount = 0;
    turn = 1;
    cx = x + 1;
    cy = y;
    for (; cx < n && cy < m; turn ? cx++ : cy++) {
      if (!matrix[cx][cy]) break;
      rightCount++;
      turn = !turn;
    }
    changeCount += (topCount + 1) * (rightCount + 1) - 1;
    if (matrix[x][y]) {
      count -= changeCount;
    } else {
      count += changeCount;
    }
    matrix[x][y] = !matrix[x][y];
    cout << count << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  solve();
}

```