#include <bits/stdc++.h>
using namespace std;
const int M = 1005;
long long read() {
  long long x = 0, f = 1;
  char c;
  while ((c = getchar()) < '0' || c > '9') {
    if (c == '-') f = -1;
  }
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + (c ^ 48);
    c = getchar();
  }
  return x * f;
}
long long n, m, q, ans, num, dp[M][M][2];
void upd(long long i, long long j, long long w) {
  if (!i || !j) return;
  if (dp[i][j][0]) {
    ans -= dp[i][j][0] + dp[i][j][1];
    dp[i][j][0] = dp[i][j][1] = 1;
    dp[i][j][0] += dp[i + 1][j][1];
    dp[i][j][1] += dp[i][j + 1][0];
    ans += dp[i][j][0] + dp[i][j][1];
  }
  if (!w)
    upd(i, j - 1, 1);
  else
    upd(i - 1, j, 0);
}
signed main() {
  n = read();
  m = read();
  q = read();
  num = n * m;
  for (long long i = n; i >= 1; i--)
    for (long long j = m; j >= 1; j--) {
      dp[i][j][0] = dp[i][j][1] = 1;
      dp[i][j][0] += dp[i + 1][j][1];
      dp[i][j][1] += dp[i][j + 1][0];
      ans += dp[i][j][0] + dp[i][j][1];
    }
  while (q--) {
    long long i = read(), j = read();
    if (dp[i][j][0]) {
      ans -= dp[i][j][0] + dp[i][j][1];
      dp[i][j][0] = dp[i][j][1] = 0;
      num--;
    } else {
      dp[i][j][0] = dp[i][j][1] = 1;
      dp[i][j][0] += dp[i + 1][j][1];
      dp[i][j][1] += dp[i][j + 1][0];
      ans += dp[i][j][0] + dp[i][j][1];
      num++;
    }
    upd(i - 1, j, 0);
    upd(i, j - 1, 1);
    printf("%lld\n", ans - num);
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 M = 1005;
long long read() {
  long long x = 0, f = 1;
  char c;
  while ((c = getchar()) < '0' || c > '9') {
    if (c == '-') f = -1;
  }
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + (c ^ 48);
    c = getchar();
  }
  return x * f;
}
long long n, m, q, ans, num, dp[M][M][2];
void upd(long long i, long long j, long long w) {
  if (!i || !j) return;
  if (dp[i][j][0]) {
    ans -= dp[i][j][0] + dp[i][j][1];
    dp[i][j][0] = dp[i][j][1] = 1;
    dp[i][j][0] += dp[i + 1][j][1];
    dp[i][j][1] += dp[i][j + 1][0];
    ans += dp[i][j][0] + dp[i][j][1];
  }
  if (!w)
    upd(i, j - 1, 1);
  else
    upd(i - 1, j, 0);
}
signed main() {
  n = read();
  m = read();
  q = read();
  num = n * m;
  for (long long i = n; i >= 1; i--)
    for (long long j = m; j >= 1; j--) {
      dp[i][j][0] = dp[i][j][1] = 1;
      dp[i][j][0] += dp[i + 1][j][1];
      dp[i][j][1] += dp[i][j + 1][0];
      ans += dp[i][j][0] + dp[i][j][1];
    }
  while (q--) {
    long long i = read(), j = read();
    if (dp[i][j][0]) {
      ans -= dp[i][j][0] + dp[i][j][1];
      dp[i][j][0] = dp[i][j][1] = 0;
      num--;
    } else {
      dp[i][j][0] = dp[i][j][1] = 1;
      dp[i][j][0] += dp[i + 1][j][1];
      dp[i][j][1] += dp[i][j + 1][0];
      ans += dp[i][j][0] + dp[i][j][1];
      num++;
    }
    upd(i - 1, j, 0);
    upd(i, j - 1, 1);
    printf("%lld\n", ans - num);
  }
}

```