#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 5;
int n, m, q, ans, f[N][N][2], g[N][N];
bool st[N][N];
int main() {
  scanf("%d %d %d", &n, &m, &q);
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      f[i][j][0] = (j == 1) ? 0 : (f[i][j - 1][1] + 1);
      f[i][j][1] = (i == 1) ? 0 : (f[i - 1][j][0] + 1);
      g[i][j] = f[i][j][0] + f[i][j][1] + 1;
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      ans += g[i][j];
    }
  }
  for (int i = 0; i <= n; ++i) {
    st[i][0] = st[i][m + 1] = 1;
  }
  for (int i = 0; i <= m; ++i) {
    st[0][i] = st[n + 1][i] = 1;
  }
  while (q--) {
    int x, y, sign;
    scanf("%d %d", &x, &y);
    sign = st[x][y] ? 1 : -1;
    int A = min(x - 1, y - 1);
    int B = min(n - x, m - y);
    int p = 0, q = 0, s = 0, t = 0;
    for (int i = 1; i <= A + 1; ++i) {
      if (st[x - i][y - i + 1]) break;
      ++p;
      if (st[x - i][y - i]) break;
      ++p;
    }
    for (int i = 1; i <= A + 1; ++i) {
      if (st[x - i + 1][y - i]) break;
      ++q;
      if (st[x - i][y - i]) break;
      ++q;
    }
    for (int i = 1; i <= B + 1; ++i) {
      if (st[x + i - 1][y + i]) break;
      ++s;
      if (st[x + i][y + i]) break;
      ++s;
    }
    for (int i = 1; i <= B + 1; ++i) {
      if (st[x + i][y + i - 1]) break;
      ++t;
      if (st[x + i][y + i]) break;
      ++t;
    }
    ans += (p * s + q * t + s + p + q + t + 1) * sign;
    st[x][y] ^= 1;
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 = 1e3 + 5;
int n, m, q, ans, f[N][N][2], g[N][N];
bool st[N][N];
int main() {
  scanf("%d %d %d", &n, &m, &q);
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      f[i][j][0] = (j == 1) ? 0 : (f[i][j - 1][1] + 1);
      f[i][j][1] = (i == 1) ? 0 : (f[i - 1][j][0] + 1);
      g[i][j] = f[i][j][0] + f[i][j][1] + 1;
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      ans += g[i][j];
    }
  }
  for (int i = 0; i <= n; ++i) {
    st[i][0] = st[i][m + 1] = 1;
  }
  for (int i = 0; i <= m; ++i) {
    st[0][i] = st[n + 1][i] = 1;
  }
  while (q--) {
    int x, y, sign;
    scanf("%d %d", &x, &y);
    sign = st[x][y] ? 1 : -1;
    int A = min(x - 1, y - 1);
    int B = min(n - x, m - y);
    int p = 0, q = 0, s = 0, t = 0;
    for (int i = 1; i <= A + 1; ++i) {
      if (st[x - i][y - i + 1]) break;
      ++p;
      if (st[x - i][y - i]) break;
      ++p;
    }
    for (int i = 1; i <= A + 1; ++i) {
      if (st[x - i + 1][y - i]) break;
      ++q;
      if (st[x - i][y - i]) break;
      ++q;
    }
    for (int i = 1; i <= B + 1; ++i) {
      if (st[x + i - 1][y + i]) break;
      ++s;
      if (st[x + i][y + i]) break;
      ++s;
    }
    for (int i = 1; i <= B + 1; ++i) {
      if (st[x + i][y + i - 1]) break;
      ++t;
      if (st[x + i][y + i]) break;
      ++t;
    }
    ans += (p * s + q * t + s + p + q + t + 1) * sign;
    st[x][y] ^= 1;
    printf("%d\n", ans);
  }
  return 0;
}

```