#include <bits/stdc++.h>
using namespace std;
const long long N = 1005;
long long n, m, q;
bool a[N][N];
long long dp1[N][N];
long long dp2[N][N];
long long dp[N][N];
long long ans = 0;
void init() {
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= m; j++) dp1[i][j] = dp2[i][j] = 1;
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= m; j++) {
      dp1[i][j] += dp2[i][j - 1];
      dp2[i][j] += dp1[i - 1][j];
      dp[i][j] = dp1[i][j] + dp2[i][j] - 1;
      ans += dp[i][j];
    }
}
void fun(long long i, long long j) {
  ans -= dp[i][j];
  if (a[i][j] == 1) {
    dp[i][j] = dp1[i][j] = dp2[i][j] = 0;
  } else {
    dp1[i][j] = dp2[i][j] = 1;
    dp1[i][j] += dp2[i][j - 1];
    dp2[i][j] += dp1[i - 1][j];
    dp[i][j] = dp1[i][j] + dp2[i][j] - 1;
  }
  ans += dp[i][j];
}
signed main() {
  cin >> n >> m >> q;
  init();
  while (q--) {
    long long x, y;
    cin >> x >> y;
    a[x][y] = !a[x][y];
    long long i = x + 1;
    long long j = y + 1;
    fun(x, y);
    while (1) {
      if (i > n || j > m) {
        if (i - 1 <= n && j <= m) fun(i - 1, j);
        if (i <= n && j - 1 <= m) fun(i, j - 1);
        break;
      }
      fun(i, j - 1);
      fun(i - 1, j);
      fun(i, j);
      i++;
      j++;
    }
    cout << ans << endl;
  }
  return 0;
}

### 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;
const long long N = 1005;
long long n, m, q;
bool a[N][N];
long long dp1[N][N];
long long dp2[N][N];
long long dp[N][N];
long long ans = 0;
void init() {
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= m; j++) dp1[i][j] = dp2[i][j] = 1;
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= m; j++) {
      dp1[i][j] += dp2[i][j - 1];
      dp2[i][j] += dp1[i - 1][j];
      dp[i][j] = dp1[i][j] + dp2[i][j] - 1;
      ans += dp[i][j];
    }
}
void fun(long long i, long long j) {
  ans -= dp[i][j];
  if (a[i][j] == 1) {
    dp[i][j] = dp1[i][j] = dp2[i][j] = 0;
  } else {
    dp1[i][j] = dp2[i][j] = 1;
    dp1[i][j] += dp2[i][j - 1];
    dp2[i][j] += dp1[i - 1][j];
    dp[i][j] = dp1[i][j] + dp2[i][j] - 1;
  }
  ans += dp[i][j];
}
signed main() {
  cin >> n >> m >> q;
  init();
  while (q--) {
    long long x, y;
    cin >> x >> y;
    a[x][y] = !a[x][y];
    long long i = x + 1;
    long long j = y + 1;
    fun(x, y);
    while (1) {
      if (i > n || j > m) {
        if (i - 1 <= n && j <= m) fun(i - 1, j);
        if (i <= n && j - 1 <= m) fun(i, j - 1);
        break;
      }
      fun(i, j - 1);
      fun(i - 1, j);
      fun(i, j);
      i++;
      j++;
    }
    cout << ans << endl;
  }
  return 0;
}

```