#include <bits/stdc++.h>
using namespace std;
void test_case() {
  int n, m, q;
  cin >> n >> m >> q;
  int64_t total = 0;
  vector<vector<pair<int64_t, int64_t>>> dp(
      n + 1, vector<pair<int64_t, int64_t>>(m + 1));
  auto cnt = [&dp](int i, int j) {
    return dp[i][j].first + dp[i][j].second - 1;
  };
  auto update = [&dp, &cnt](int i, int j) {
    dp[i][j].second = dp[i - 1][j].first + 1;
    dp[i][j].first = dp[i][j - 1].second + 1;
    return cnt(i, j);
  };
  auto locked = [&dp](int i, int j) {
    return dp[i][j].first == 0 && dp[i][j].second == 0;
  };
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      update(i, j);
      total += cnt(i, j);
    }
  }
  while (q--) {
    int i, j;
    cin >> i >> j;
    if (!locked(i, j)) {
      total -= cnt(i, j);
      dp[i][j] = make_pair(0, 0);
    } else
      total += update(i, j);
    for (++i, ++j; i - 1 <= n && j - 1 <= m; ++i, ++j) {
      if (j <= m && !locked(i - 1, j)) {
        total -= cnt(i - 1, j);
        total += update(i - 1, j);
      }
      if (i <= n && !locked(i, j - 1)) {
        total -= cnt(i, j - 1);
        total += update(i, j - 1);
      }
      if (i <= n && j <= m && !locked(i, j)) {
        total -= cnt(i, j);
        total += update(i, j);
      }
    }
    cout << total << '\n';
  }
}
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int Case = 1;
  for (int i = 1; i <= Case; i++) {
    test_case();
  }
  return 0;
}

### Prompt

Generate 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;
void test_case() {
  int n, m, q;
  cin >> n >> m >> q;
  int64_t total = 0;
  vector<vector<pair<int64_t, int64_t>>> dp(
      n + 1, vector<pair<int64_t, int64_t>>(m + 1));
  auto cnt = [&dp](int i, int j) {
    return dp[i][j].first + dp[i][j].second - 1;
  };
  auto update = [&dp, &cnt](int i, int j) {
    dp[i][j].second = dp[i - 1][j].first + 1;
    dp[i][j].first = dp[i][j - 1].second + 1;
    return cnt(i, j);
  };
  auto locked = [&dp](int i, int j) {
    return dp[i][j].first == 0 && dp[i][j].second == 0;
  };
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      update(i, j);
      total += cnt(i, j);
    }
  }
  while (q--) {
    int i, j;
    cin >> i >> j;
    if (!locked(i, j)) {
      total -= cnt(i, j);
      dp[i][j] = make_pair(0, 0);
    } else
      total += update(i, j);
    for (++i, ++j; i - 1 <= n && j - 1 <= m; ++i, ++j) {
      if (j <= m && !locked(i - 1, j)) {
        total -= cnt(i - 1, j);
        total += update(i - 1, j);
      }
      if (i <= n && !locked(i, j - 1)) {
        total -= cnt(i, j - 1);
        total += update(i, j - 1);
      }
      if (i <= n && j <= m && !locked(i, j)) {
        total -= cnt(i, j);
        total += update(i, j);
      }
    }
    cout << total << '\n';
  }
}
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int Case = 1;
  for (int i = 1; i <= Case; i++) {
    test_case();
  }
  return 0;
}

```