#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000 + 5;
int n, m, q, t;
long long f[maxn][maxn][2];
long long ans = 0;
bool vis[maxn][maxn];
long long up(int x, int y, int d) {
  if (x < 1 or y < 1) return 0;
  if (vis[x][y]) return 0;
  if (d == 0) {
    return up(x - 1, y, 1) + 1;
  } else {
    return up(x, y - 1, 0) + 1;
  }
}
long long down(int x, int y, int d) {
  if (x > n or y > m) return 0;
  if (vis[x][y]) return 0;
  if (d == 0) {
    return down(x + 1, y, 1) + 1;
  } else {
    return down(x, y + 1, 0) + 1;
  }
}
int main() {
  int i, j;
  scanf("%d%d%d", &n, &m, &q);
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= m; j++) {
      f[i][j][0] = 1;
      f[i][j][1] = 1;
      f[i][j][1] += f[i][j - 1][0];
      f[i][j][0] += f[i - 1][j][1];
      ans += f[i][j][0] + f[i][j][1] - 1;
    }
  }
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (vis[x][y]) {
      vis[x][y] = false;
      ans += up(x, y, 0) * down(x, y, 1) + up(x, y, 1) * down(x, y, 0) - 1;
    } else {
      long long a = +-1;
      long long b = +-1;
      ans -= up(x, y, 0) * down(x, y, 1) + up(x, y, 1) * down(x, y, 0) - 1;
      vis[x][y] = true;
    }
    printf("%lld\n", ans);
  }
  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 maxn = 1000 + 5;
int n, m, q, t;
long long f[maxn][maxn][2];
long long ans = 0;
bool vis[maxn][maxn];
long long up(int x, int y, int d) {
  if (x < 1 or y < 1) return 0;
  if (vis[x][y]) return 0;
  if (d == 0) {
    return up(x - 1, y, 1) + 1;
  } else {
    return up(x, y - 1, 0) + 1;
  }
}
long long down(int x, int y, int d) {
  if (x > n or y > m) return 0;
  if (vis[x][y]) return 0;
  if (d == 0) {
    return down(x + 1, y, 1) + 1;
  } else {
    return down(x, y + 1, 0) + 1;
  }
}
int main() {
  int i, j;
  scanf("%d%d%d", &n, &m, &q);
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= m; j++) {
      f[i][j][0] = 1;
      f[i][j][1] = 1;
      f[i][j][1] += f[i][j - 1][0];
      f[i][j][0] += f[i - 1][j][1];
      ans += f[i][j][0] + f[i][j][1] - 1;
    }
  }
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (vis[x][y]) {
      vis[x][y] = false;
      ans += up(x, y, 0) * down(x, y, 1) + up(x, y, 1) * down(x, y, 0) - 1;
    } else {
      long long a = +-1;
      long long b = +-1;
      ans -= up(x, y, 0) * down(x, y, 1) + up(x, y, 1) * down(x, y, 0) - 1;
      vis[x][y] = true;
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```