#include <bits/stdc++.h>
using namespace std;
long long dp[1005][1005][2];
bool ban[1005][1005];
void solve() {
  int n, m, q;
  cin >> n >> m >> q;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      dp[i][j][0] = dp[i - 1][j][1] + 1;
      dp[i][j][1] = dp[i][j - 1][0] + 1;
      ans += (dp[i][j][0] + dp[i][j][1]);
    }
  }
  int open = n * m;
  while (q--) {
    int x, y;
    cin >> x >> y;
    if (!ban[x][y]) {
      long long a = dp[x][y][0];
      long long b = dp[x][y][1];
      dp[x][y][0] = 0;
      dp[x][y][1] = 0;
      open--;
      ans -= (a + b);
      bool last = 1;
      int tmp = x, tmp2 = y;
      while (x <= n && y <= m && !ban[x][y]) {
        if (!last) {
          x++;
        } else
          y++;
        if (x <= n && y <= m && !ban[x][y]) {
          dp[x][y][last] -= a;
          ans -= a;
        }
        last = !last;
      }
      x = tmp, y = tmp2;
      last = 0;
      while (x <= n && y <= m && !ban[x][y]) {
        if (!last) {
          x++;
        } else
          y++;
        if (x <= n && y <= m && !ban[x][y]) {
          dp[x][y][last] -= b;
          ans -= b;
        }
        last = !last;
      }
      ban[tmp][tmp2] = 1;
    } else {
      long long a = dp[x - 1][y][1] + 1;
      long long b = dp[x][y - 1][0] + 1;
      dp[x][y][0] = a;
      dp[x][y][1] = b;
      open++;
      ans += (a + b);
      bool last = 1;
      int tmp = x, tmp2 = y;
      ban[x][y] = 0;
      while (x <= n && y <= m && !ban[x][y]) {
        if (!last) {
          x++;
        } else
          y++;
        if (x <= n && y <= m && !ban[x][y]) {
          dp[x][y][last] += a;
          ans += a;
        }
        last = !last;
      }
      x = tmp, y = tmp2;
      last = 0;
      while (x <= n && y <= m && !ban[x][y]) {
        if (!last) {
          x++;
        } else
          y++;
        if (x <= n && y <= m && !ban[x][y]) {
          dp[x][y][last] += b;
          ans += b;
        }
        last = !last;
      }
    }
    cout << ans - open << "\n";
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t = 1;
  while (t--) {
    solve();
  }
}

### Prompt

Please create a solution in cpp 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;
long long dp[1005][1005][2];
bool ban[1005][1005];
void solve() {
  int n, m, q;
  cin >> n >> m >> q;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      dp[i][j][0] = dp[i - 1][j][1] + 1;
      dp[i][j][1] = dp[i][j - 1][0] + 1;
      ans += (dp[i][j][0] + dp[i][j][1]);
    }
  }
  int open = n * m;
  while (q--) {
    int x, y;
    cin >> x >> y;
    if (!ban[x][y]) {
      long long a = dp[x][y][0];
      long long b = dp[x][y][1];
      dp[x][y][0] = 0;
      dp[x][y][1] = 0;
      open--;
      ans -= (a + b);
      bool last = 1;
      int tmp = x, tmp2 = y;
      while (x <= n && y <= m && !ban[x][y]) {
        if (!last) {
          x++;
        } else
          y++;
        if (x <= n && y <= m && !ban[x][y]) {
          dp[x][y][last] -= a;
          ans -= a;
        }
        last = !last;
      }
      x = tmp, y = tmp2;
      last = 0;
      while (x <= n && y <= m && !ban[x][y]) {
        if (!last) {
          x++;
        } else
          y++;
        if (x <= n && y <= m && !ban[x][y]) {
          dp[x][y][last] -= b;
          ans -= b;
        }
        last = !last;
      }
      ban[tmp][tmp2] = 1;
    } else {
      long long a = dp[x - 1][y][1] + 1;
      long long b = dp[x][y - 1][0] + 1;
      dp[x][y][0] = a;
      dp[x][y][1] = b;
      open++;
      ans += (a + b);
      bool last = 1;
      int tmp = x, tmp2 = y;
      ban[x][y] = 0;
      while (x <= n && y <= m && !ban[x][y]) {
        if (!last) {
          x++;
        } else
          y++;
        if (x <= n && y <= m && !ban[x][y]) {
          dp[x][y][last] += a;
          ans += a;
        }
        last = !last;
      }
      x = tmp, y = tmp2;
      last = 0;
      while (x <= n && y <= m && !ban[x][y]) {
        if (!last) {
          x++;
        } else
          y++;
        if (x <= n && y <= m && !ban[x][y]) {
          dp[x][y][last] += b;
          ans += b;
        }
        last = !last;
      }
    }
    cout << ans - open << "\n";
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t = 1;
  while (t--) {
    solve();
  }
}

```