#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n, m;
  cin >> n >> m;
  vector<vector<vector<int>>> dp(n + 2,
                                 vector<vector<int>>(m + 2, vector<int>(4, 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;
    }
  for (int i = n; i >= 1; i--)
    for (int j = m; j >= 1; j--) {
      dp[i][j][2] = dp[i + 1][j][3] + 1;
      dp[i][j][3] = dp[i][j + 1][2] + 1;
    }
  int ans = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      ans += (dp[i][j][0]);
      ans += (dp[i][j][1]);
    }
  ans -= n * m;
  int q;
  cin >> q;
  vector<vector<int>> vis(n + 2, vector<int>(m + 2, 0));
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      vis[i][j] = 1;
    }
  for (int i = 0; i < q; i++) {
    int x, y;
    cin >> x >> y;
    if (vis[x][y] == 1) {
      vis[x][y] ^= 1;
      ans -= (dp[x][y - 1][0] + 1) * (dp[x + 1][y][3] + 1);
      ans -= (dp[x][y + 1][2] + 1) * (dp[x - 1][y][1] + 1);
      ans++;
      for (int k = 0; k < 4; k++) dp[x][y][k] = 0;
    } else {
      vis[x][y] ^= 1;
      ans += (dp[x][y - 1][0] + 1) * (dp[x + 1][y][3] + 1);
      ans += (dp[x][y + 1][2] + 1) * (dp[x - 1][y][1] + 1);
      ans--;
      for (int k = 0; k < 4; k++) dp[x][y][k] = 1;
    }
    int px = x, py = y;
    int qx = x, qy = y;
    for (int k = 0; k <= 2 * max(n, m); k++) {
      if (k & 1) {
        py++;
        qx++;
      } else if (k != 0) {
        px++;
        qy++;
      }
      if (px <= n && py <= m && vis[px][py])
        dp[px][py][1] = 1 + dp[px][py - 1][0];
      if (qx <= n && qy <= m && vis[qx][qy])
        dp[qx][qy][0] = 1 + dp[qx - 1][qy][1];
    }
    px = x, py = y;
    qx = x, qy = y;
    for (int k = 0; k <= 2 * max(n, m); k++) {
      if (k & 1) {
        py--;
        qx--;
      } else if (k != 0) {
        px--;
        qy--;
      }
      if (px >= 1 && py >= 1 && vis[px][py])
        dp[px][py][3] = 1 + dp[px][py + 1][2];
      if (qx >= 1 && qy >= 1 && vis[qx][qy])
        dp[qx][qy][2] = 1 + dp[qx + 1][qy][3];
    }
    cout << ans << endl;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int T = 1;
  while (T--) {
    solve();
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 + 2,
                                 vector<vector<int>>(m + 2, vector<int>(4, 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;
    }
  for (int i = n; i >= 1; i--)
    for (int j = m; j >= 1; j--) {
      dp[i][j][2] = dp[i + 1][j][3] + 1;
      dp[i][j][3] = dp[i][j + 1][2] + 1;
    }
  int ans = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      ans += (dp[i][j][0]);
      ans += (dp[i][j][1]);
    }
  ans -= n * m;
  int q;
  cin >> q;
  vector<vector<int>> vis(n + 2, vector<int>(m + 2, 0));
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      vis[i][j] = 1;
    }
  for (int i = 0; i < q; i++) {
    int x, y;
    cin >> x >> y;
    if (vis[x][y] == 1) {
      vis[x][y] ^= 1;
      ans -= (dp[x][y - 1][0] + 1) * (dp[x + 1][y][3] + 1);
      ans -= (dp[x][y + 1][2] + 1) * (dp[x - 1][y][1] + 1);
      ans++;
      for (int k = 0; k < 4; k++) dp[x][y][k] = 0;
    } else {
      vis[x][y] ^= 1;
      ans += (dp[x][y - 1][0] + 1) * (dp[x + 1][y][3] + 1);
      ans += (dp[x][y + 1][2] + 1) * (dp[x - 1][y][1] + 1);
      ans--;
      for (int k = 0; k < 4; k++) dp[x][y][k] = 1;
    }
    int px = x, py = y;
    int qx = x, qy = y;
    for (int k = 0; k <= 2 * max(n, m); k++) {
      if (k & 1) {
        py++;
        qx++;
      } else if (k != 0) {
        px++;
        qy++;
      }
      if (px <= n && py <= m && vis[px][py])
        dp[px][py][1] = 1 + dp[px][py - 1][0];
      if (qx <= n && qy <= m && vis[qx][qy])
        dp[qx][qy][0] = 1 + dp[qx - 1][qy][1];
    }
    px = x, py = y;
    qx = x, qy = y;
    for (int k = 0; k <= 2 * max(n, m); k++) {
      if (k & 1) {
        py--;
        qx--;
      } else if (k != 0) {
        px--;
        qy--;
      }
      if (px >= 1 && py >= 1 && vis[px][py])
        dp[px][py][3] = 1 + dp[px][py + 1][2];
      if (qx >= 1 && qy >= 1 && vis[qx][qy])
        dp[qx][qy][2] = 1 + dp[qx + 1][qy][3];
    }
    cout << ans << endl;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int T = 1;
  while (T--) {
    solve();
  }
}

```