#include <bits/stdc++.h>
using namespace std;
const long long N = 1005;
long long n, m, q, ans;
bool vizitat[N][N];
long long dp[N][N][2];
bool verifica(long long i, long long j) {
  if (i < 1 || j < 1 || i > n || j > m) return false;
  return true;
}
long long cnt(long long i, long long j) {
  long long answer = 0;
  for (long long it = 0; it <= 1; it++) {
    long long up = 0;
    long long lin = i;
    long long col = j;
    while (verifica(lin, col) && vizitat[lin][col] == false) {
      if (up % 2 == it) {
        lin--;
      } else {
        col--;
      }
      up++;
    }
    long long down = 0;
    lin = i;
    col = j;
    while (verifica(lin, col) && vizitat[lin][col] == false) {
      if (down % 2 == it) {
        col++;
      } else {
        lin++;
      }
      down++;
    }
    answer += up * down;
  }
  answer--;
  return answer;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  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++) {
      dp[i][j][1] = dp[i][j - 1][0] + 1;
      dp[i][j][0] = dp[i - 1][j][1] + 1;
      ans += dp[i][j][1];
      ans += dp[i][j][0];
    }
  }
  ans -= n * m;
  while (q--) {
    long long lin, col;
    cin >> lin >> col;
    if (vizitat[lin][col] == true) {
      vizitat[lin][col] = false;
      ans += cnt(lin, col);
    } else {
      ans -= cnt(lin, col);
      vizitat[lin][col] = true;
    }
    cout << ans << '\n';
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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, ans;
bool vizitat[N][N];
long long dp[N][N][2];
bool verifica(long long i, long long j) {
  if (i < 1 || j < 1 || i > n || j > m) return false;
  return true;
}
long long cnt(long long i, long long j) {
  long long answer = 0;
  for (long long it = 0; it <= 1; it++) {
    long long up = 0;
    long long lin = i;
    long long col = j;
    while (verifica(lin, col) && vizitat[lin][col] == false) {
      if (up % 2 == it) {
        lin--;
      } else {
        col--;
      }
      up++;
    }
    long long down = 0;
    lin = i;
    col = j;
    while (verifica(lin, col) && vizitat[lin][col] == false) {
      if (down % 2 == it) {
        col++;
      } else {
        lin++;
      }
      down++;
    }
    answer += up * down;
  }
  answer--;
  return answer;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  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++) {
      dp[i][j][1] = dp[i][j - 1][0] + 1;
      dp[i][j][0] = dp[i - 1][j][1] + 1;
      ans += dp[i][j][1];
      ans += dp[i][j][0];
    }
  }
  ans -= n * m;
  while (q--) {
    long long lin, col;
    cin >> lin >> col;
    if (vizitat[lin][col] == true) {
      vizitat[lin][col] = false;
      ans += cnt(lin, col);
    } else {
      ans -= cnt(lin, col);
      vizitat[lin][col] = true;
    }
    cout << ans << '\n';
  }
  return 0;
}

```