#include <bits/stdc++.h>
using namespace std;
long long n, m, k, cnt, ans;
long long dp[1010][1010][2], x, y;
bool ma[1010][1010];
void change(long long x, long long y) {
  if (ma[x][y]) return;
  if (x > n || y > m) return;
  ans -= dp[x][y][1] + dp[x][y][0];
  dp[x][y][1] = dp[x - 1][y][0] + 1;
  dp[x][y][0] = dp[x][y - 1][1] + 1;
  ans += dp[x][y][1] + dp[x][y][0];
}
signed main() {
  scanf("%lld%lld%lld", &n, &m, &k);
  cnt = n * m;
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= m; j++) {
      dp[i][j][1] = dp[i - 1][j][0] + 1;
      dp[i][j][0] = dp[i][j - 1][1] + 1;
    }
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= m; j++) ans += (dp[i][j][0] + dp[i][j][1]);
  while (k--) {
    scanf("%lld%lld", &x, &y);
    if (!ma[x][y]) {
      cnt--;
      ans -= dp[x][y][0] + dp[x][y][1];
      dp[x][y][0] = dp[x][y][1] = 0;
      ma[x][y] = 1;
    } else {
      cnt++;
      dp[x][y][0] = dp[x][y - 1][1] + 1;
      dp[x][y][1] = dp[x - 1][y][0] + 1;
      ans += dp[x][y][0] + dp[x][y][1];
      ma[x][y] = 0;
    }
    while (x <= n && y <= m) {
      change(x + 1, y);
      change(x, y + 1);
      change(x + 1, y + 1);
      x++;
      y++;
    }
    printf("%lld\n", ans - cnt);
  }
}

### Prompt

Your challenge is to write 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;
long long n, m, k, cnt, ans;
long long dp[1010][1010][2], x, y;
bool ma[1010][1010];
void change(long long x, long long y) {
  if (ma[x][y]) return;
  if (x > n || y > m) return;
  ans -= dp[x][y][1] + dp[x][y][0];
  dp[x][y][1] = dp[x - 1][y][0] + 1;
  dp[x][y][0] = dp[x][y - 1][1] + 1;
  ans += dp[x][y][1] + dp[x][y][0];
}
signed main() {
  scanf("%lld%lld%lld", &n, &m, &k);
  cnt = n * m;
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= m; j++) {
      dp[i][j][1] = dp[i - 1][j][0] + 1;
      dp[i][j][0] = dp[i][j - 1][1] + 1;
    }
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= m; j++) ans += (dp[i][j][0] + dp[i][j][1]);
  while (k--) {
    scanf("%lld%lld", &x, &y);
    if (!ma[x][y]) {
      cnt--;
      ans -= dp[x][y][0] + dp[x][y][1];
      dp[x][y][0] = dp[x][y][1] = 0;
      ma[x][y] = 1;
    } else {
      cnt++;
      dp[x][y][0] = dp[x][y - 1][1] + 1;
      dp[x][y][1] = dp[x - 1][y][0] + 1;
      ans += dp[x][y][0] + dp[x][y][1];
      ma[x][y] = 0;
    }
    while (x <= n && y <= m) {
      change(x + 1, y);
      change(x, y + 1);
      change(x + 1, y + 1);
      x++;
      y++;
    }
    printf("%lld\n", ans - cnt);
  }
}

```