#include <bits/stdc++.h>
using namespace std;
const long long N = 1e3 + 7;
long long n;
long long m;
long long dp[N][N][2];
long long on[N][N];
long long all;
long long q;
void calc(long long x, long long y) {
  long long ban = on[x][y] * (dp[x][y][0] + dp[x][y][1] - 1);
  all += ban;
  for (long long i = 0; i <= 1; i++) {
    long long a = x, b = y;
    bool c = i;
    while (1) {
      if (c == 1)
        ++b;
      else
        ++a;
      if (a > n || b > m) break;
      c ^= 1;
      dp[a][b][c] += on[x][y] * dp[x][y][i];
      if (on[a][b] == 1) break;
      all += on[x][y] * dp[x][y][i];
    }
  }
  on[x][y] = -on[x][y];
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> q;
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= m; j++) {
      on[i][j] = -1;
      dp[i][j][1] = dp[i - 1][j][0] + 1;
      dp[i][j][0] = dp[i][j - 1][1] + 1;
      all += dp[i][j][0] + dp[i][j][1] - 1;
    }
  }
  while (q--) {
    long long x, y;
    cin >> x >> y;
    calc(x, y);
    cout << all << '\n';
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 = 1e3 + 7;
long long n;
long long m;
long long dp[N][N][2];
long long on[N][N];
long long all;
long long q;
void calc(long long x, long long y) {
  long long ban = on[x][y] * (dp[x][y][0] + dp[x][y][1] - 1);
  all += ban;
  for (long long i = 0; i <= 1; i++) {
    long long a = x, b = y;
    bool c = i;
    while (1) {
      if (c == 1)
        ++b;
      else
        ++a;
      if (a > n || b > m) break;
      c ^= 1;
      dp[a][b][c] += on[x][y] * dp[x][y][i];
      if (on[a][b] == 1) break;
      all += on[x][y] * dp[x][y][i];
    }
  }
  on[x][y] = -on[x][y];
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> q;
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= m; j++) {
      on[i][j] = -1;
      dp[i][j][1] = dp[i - 1][j][0] + 1;
      dp[i][j][0] = dp[i][j - 1][1] + 1;
      all += dp[i][j][0] + dp[i][j][1] - 1;
    }
  }
  while (q--) {
    long long x, y;
    cin >> x >> y;
    calc(x, y);
    cout << all << '\n';
  }
}

```