#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
int m, n, q;
long long f[N][N][2];
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m >> q;
  memset(f, 0, sizeof f);
  long long cnt = n * m;
  long long total = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      f[i][j][0] = 1;
      f[i][j][1] = 1;
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    for (int j = m - 1; j >= 0; j--) {
      f[i][j][0] += f[i][j + 1][1];
      f[i][j][1] += f[i + 1][j][0];
      total += (f[i][j][0] + f[i][j][1]);
    }
  }
  for (int i = 0; i < q; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    int change0, change1;
    if (f[x][y][0] > 0) {
      change0 = -f[x][y][1];
      change1 = -f[x][y][0];
      f[x][y][0] = f[x][y][1] = 0;
      total += (change0 + change1);
      cnt--;
    } else {
      change0 = f[x][y][1] = f[x + 1][y][0] + 1;
      change1 = f[x][y][0] = f[x][y + 1][1] + 1;
      total += (change0 + change1);
      cnt++;
    }
    int t = 1;
    int xx = x;
    int yy = y;
    bool failed = false;
    while (!failed) {
      if (t & 1) {
        x--;
        if (x < 0 || y < 0 || f[x][y][0] == 0)
          failed = true;
        else {
          f[x][y][1] += change1;
          total += change1;
        }
      } else {
        y--;
        if (x < 0 || y < 0 || f[x][y][0] == 0)
          failed = true;
        else {
          f[x][y][0] += change1;
          total += change1;
        }
      }
      t++;
    }
    x = xx;
    y = yy;
    t = 1;
    failed = false;
    while (!failed) {
      if (t & 1) {
        y--;
        if (x < 0 || y < 0 or f[x][y][0] == 0)
          failed = true;
        else {
          f[x][y][0] += change0;
          total += change0;
        }
      } else {
        x--;
        if (x < 0 || y < 0 or f[x][y][0] == 0)
          failed = true;
        else {
          f[x][y][1] += change0;
          total += change0;
        }
      }
      t++;
    }
    cout << total - cnt << endl;
  }
  return 0;
}

### 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;
const int N = 1010;
int m, n, q;
long long f[N][N][2];
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m >> q;
  memset(f, 0, sizeof f);
  long long cnt = n * m;
  long long total = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      f[i][j][0] = 1;
      f[i][j][1] = 1;
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    for (int j = m - 1; j >= 0; j--) {
      f[i][j][0] += f[i][j + 1][1];
      f[i][j][1] += f[i + 1][j][0];
      total += (f[i][j][0] + f[i][j][1]);
    }
  }
  for (int i = 0; i < q; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    int change0, change1;
    if (f[x][y][0] > 0) {
      change0 = -f[x][y][1];
      change1 = -f[x][y][0];
      f[x][y][0] = f[x][y][1] = 0;
      total += (change0 + change1);
      cnt--;
    } else {
      change0 = f[x][y][1] = f[x + 1][y][0] + 1;
      change1 = f[x][y][0] = f[x][y + 1][1] + 1;
      total += (change0 + change1);
      cnt++;
    }
    int t = 1;
    int xx = x;
    int yy = y;
    bool failed = false;
    while (!failed) {
      if (t & 1) {
        x--;
        if (x < 0 || y < 0 || f[x][y][0] == 0)
          failed = true;
        else {
          f[x][y][1] += change1;
          total += change1;
        }
      } else {
        y--;
        if (x < 0 || y < 0 || f[x][y][0] == 0)
          failed = true;
        else {
          f[x][y][0] += change1;
          total += change1;
        }
      }
      t++;
    }
    x = xx;
    y = yy;
    t = 1;
    failed = false;
    while (!failed) {
      if (t & 1) {
        y--;
        if (x < 0 || y < 0 or f[x][y][0] == 0)
          failed = true;
        else {
          f[x][y][0] += change0;
          total += change0;
        }
      } else {
        x--;
        if (x < 0 || y < 0 or f[x][y][0] == 0)
          failed = true;
        else {
          f[x][y][1] += change0;
          total += change0;
        }
      }
      t++;
    }
    cout << total - cnt << endl;
  }
  return 0;
}

```