#include <bits/stdc++.h>
using namespace std;
const int N = 1010, mod = 1e9 + 7;
int arr[N][N];
long long dp1[N][N], dp2[N][N];
void print(long long dp[][N], int n, int m) {
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) cout << dp[i + 1][j + 1] << ' ';
    cout << endl;
  }
  cout << "Debug" << endl;
}
void solve() {
  int n, m, k, x, y;
  cin >> n >> m >> k;
  dp1[n][m] = dp2[n][m] = 0;
  long long ans = 0;
  for (int i = n; i > 0; i--) {
    for (int j = m; j > 0; j--) {
      arr[i][j] = 0;
      dp1[i][j] = j < m ? 1 + dp2[i][j + 1] : 0;
      dp2[i][j] = i < n ? 1 + dp1[i + 1][j] : 0;
      ans += 1 + dp1[i][j] + dp2[i][j];
    }
  }
  while (k--) {
    cin >> x >> y;
    arr[x][y] ^= 1;
    if (arr[x][y]) {
      ans = ans - 1 - dp1[x][y] - dp2[x][y];
      dp1[x][y] = dp2[x][y] = 0;
      if (x > 1) {
        ans -= dp2[x - 1][y];
        dp2[x - 1][y] = 0;
      }
      if (y > 1) {
        ans -= dp1[x][y - 1];
        dp1[x][y - 1] = 0;
      }
      x -= 1, y -= 1;
    } else
      ans++;
    for (int i = x, j = y, k = 1; i > 0 and j > 0 and !arr[i][j]; k = 3 - k) {
      if (k == 1) {
        ans -= dp1[i][j];
        dp1[i][j] = (j < m and !arr[i][j + 1]) ? 1 + dp2[i][j + 1] : 0;
        ans += dp1[i][j];
        i -= 1;
      } else {
        ans -= dp2[i][j];
        dp2[i][j] = (i < n and !arr[i + 1][j]) ? 1 + dp1[i + 1][j] : 0;
        ans += dp2[i][j];
        j -= 1;
      }
    }
    for (int i = x, j = y, k = 2; i > 0 and j > 0 and !arr[i][j]; k = 3 - k) {
      if (k == 1) {
        ans -= dp1[i][j];
        dp1[i][j] = (j < m and !arr[i][j + 1]) ? 1 + dp2[i][j + 1] : 0;
        ans += dp1[i][j];
        i -= 1;
      } else {
        ans -= dp2[i][j];
        dp2[i][j] = (i < n and !arr[i + 1][j]) ? 1 + dp1[i + 1][j] : 0;
        ans += dp2[i][j];
        j -= 1;
      }
    }
    cout << ans << '\n';
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}

### Prompt

Your task is to create 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;
const int N = 1010, mod = 1e9 + 7;
int arr[N][N];
long long dp1[N][N], dp2[N][N];
void print(long long dp[][N], int n, int m) {
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) cout << dp[i + 1][j + 1] << ' ';
    cout << endl;
  }
  cout << "Debug" << endl;
}
void solve() {
  int n, m, k, x, y;
  cin >> n >> m >> k;
  dp1[n][m] = dp2[n][m] = 0;
  long long ans = 0;
  for (int i = n; i > 0; i--) {
    for (int j = m; j > 0; j--) {
      arr[i][j] = 0;
      dp1[i][j] = j < m ? 1 + dp2[i][j + 1] : 0;
      dp2[i][j] = i < n ? 1 + dp1[i + 1][j] : 0;
      ans += 1 + dp1[i][j] + dp2[i][j];
    }
  }
  while (k--) {
    cin >> x >> y;
    arr[x][y] ^= 1;
    if (arr[x][y]) {
      ans = ans - 1 - dp1[x][y] - dp2[x][y];
      dp1[x][y] = dp2[x][y] = 0;
      if (x > 1) {
        ans -= dp2[x - 1][y];
        dp2[x - 1][y] = 0;
      }
      if (y > 1) {
        ans -= dp1[x][y - 1];
        dp1[x][y - 1] = 0;
      }
      x -= 1, y -= 1;
    } else
      ans++;
    for (int i = x, j = y, k = 1; i > 0 and j > 0 and !arr[i][j]; k = 3 - k) {
      if (k == 1) {
        ans -= dp1[i][j];
        dp1[i][j] = (j < m and !arr[i][j + 1]) ? 1 + dp2[i][j + 1] : 0;
        ans += dp1[i][j];
        i -= 1;
      } else {
        ans -= dp2[i][j];
        dp2[i][j] = (i < n and !arr[i + 1][j]) ? 1 + dp1[i + 1][j] : 0;
        ans += dp2[i][j];
        j -= 1;
      }
    }
    for (int i = x, j = y, k = 2; i > 0 and j > 0 and !arr[i][j]; k = 3 - k) {
      if (k == 1) {
        ans -= dp1[i][j];
        dp1[i][j] = (j < m and !arr[i][j + 1]) ? 1 + dp2[i][j + 1] : 0;
        ans += dp1[i][j];
        i -= 1;
      } else {
        ans -= dp2[i][j];
        dp2[i][j] = (i < n and !arr[i + 1][j]) ? 1 + dp1[i + 1][j] : 0;
        ans += dp2[i][j];
        j -= 1;
      }
    }
    cout << ans << '\n';
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}

```