#include <bits/stdc++.h>
using namespace std;
const long long md = 1000000007;
long long icase = 1;
void google(long long t) { cout << "Case #" << t << ": "; }
const long long limt = 1e3 + 5;
long long ans;
long long dp[2][limt][limt];
bool state[limt][limt];
long long m, n, k;
void update(long long r, long long c) {
  if (r > m || c > n) return;
  ans -= dp[0][r][c] + dp[1][r][c];
  if (state[r][c]) {
    dp[0][r][c] = dp[1][r][c] = 0;
  } else {
    dp[0][r][c] = dp[1][r][c - 1] + 1;
    dp[1][r][c] = dp[0][r - 1][c] + 1;
    ans += dp[0][r][c] + dp[1][r][c];
  }
}
void subMain() {
  cin >> m >> n >> k;
  for (long long i = 1; i <= m; i++) {
    for (long long j = 1; j <= n; j++) {
      update(i, j);
    }
  }
  long long r, c;
  ans -= m * n;
  while (k--) {
    cin >> r >> c;
    state[r][c] ^= 1;
    ans += (state[r][c] ? 1 : -1);
    for (long long i = r; i <= m; i++) {
      update(i, c + i - r);
      update(i, c + i - r + 1);
      update(i + 1, c + i - r);
    }
    cout << ans << endl;
  }
}
int32_t main() {
  long long t = 1;
  while (t--) {
    subMain();
  }
  return 0;
}

### Prompt

Your task is to create 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;
const long long md = 1000000007;
long long icase = 1;
void google(long long t) { cout << "Case #" << t << ": "; }
const long long limt = 1e3 + 5;
long long ans;
long long dp[2][limt][limt];
bool state[limt][limt];
long long m, n, k;
void update(long long r, long long c) {
  if (r > m || c > n) return;
  ans -= dp[0][r][c] + dp[1][r][c];
  if (state[r][c]) {
    dp[0][r][c] = dp[1][r][c] = 0;
  } else {
    dp[0][r][c] = dp[1][r][c - 1] + 1;
    dp[1][r][c] = dp[0][r - 1][c] + 1;
    ans += dp[0][r][c] + dp[1][r][c];
  }
}
void subMain() {
  cin >> m >> n >> k;
  for (long long i = 1; i <= m; i++) {
    for (long long j = 1; j <= n; j++) {
      update(i, j);
    }
  }
  long long r, c;
  ans -= m * n;
  while (k--) {
    cin >> r >> c;
    state[r][c] ^= 1;
    ans += (state[r][c] ? 1 : -1);
    for (long long i = r; i <= m; i++) {
      update(i, c + i - r);
      update(i, c + i - r + 1);
      update(i + 1, c + i - r);
    }
    cout << ans << endl;
  }
}
int32_t main() {
  long long t = 1;
  while (t--) {
    subMain();
  }
  return 0;
}

```