#include <bits/stdc++.h>
using namespace std;
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(ios::badbit | ios::failbit);
  int n, m, qn;
  cin >> n >> m >> qn;
  vector<vector<int>> state(n, vector<int>(m));
  vector<set<int>> notfree(n + m + 10);
  int delta = m + 5;
  for (auto i = -2; i <= n + 2; ++i) {
    for (auto j = -2; j <= m + 2; ++j) {
      notfree[i - j + delta].insert(i);
    }
  }
  42;
  long long res = 0;
  auto calc = [&](int i, int j) -> long long {
    42;
    42;
    42;
    42;
    42;
    42;
    long long res = -1;
    auto temp = notfree[i - j + delta].lower_bound(i);
    int x = *temp - i;
    int rx = i - *prev(temp);
    42;
    {
      auto it = notfree[i - j - 1 + delta].lower_bound(i);
      42;
      int y = *it - i;
      int p = x <= y ? 2 * x : 2 * y + 1;
      int ry = i - 1 - *prev(it);
      int q = rx <= ry ? 2 * rx : 2 * ry + 1;
      42;
      res += 1LL * p * q;
    }
    {
      auto it = notfree[i - j + 1 + delta].lower_bound(i + 1);
      42;
      int y = *it - i - 1;
      int p = x <= y ? 2 * x : 2 * y + 1;
      int ry = i - *prev(it);
      int q = rx <= ry ? 2 * rx : 2 * ry + 1;
      42;
      res += 1LL * p * q;
    }
    42;
    return res;
  };
  auto flip = [&](int i, int j) -> void {
    if (state[i][j]) {
      res -= calc(i, j);
      state[i][j] ^= 1;
      notfree[i - j + delta].insert(i);
    } else {
      state[i][j] ^= 1;
      notfree[i - j + delta].erase(i);
      res += calc(i, j);
    }
  };
  for (auto i = 0; i < n; ++i) {
    for (auto j = 0; j < m; ++j) {
      flip(i, j);
    }
  }
  42;
  for (auto qi = 0; qi < qn; ++qi) {
    int i, j;
    cin >> i >> j, --i, --j;
    flip(i, j);
    cout << res << "\n";
  }
  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;
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(ios::badbit | ios::failbit);
  int n, m, qn;
  cin >> n >> m >> qn;
  vector<vector<int>> state(n, vector<int>(m));
  vector<set<int>> notfree(n + m + 10);
  int delta = m + 5;
  for (auto i = -2; i <= n + 2; ++i) {
    for (auto j = -2; j <= m + 2; ++j) {
      notfree[i - j + delta].insert(i);
    }
  }
  42;
  long long res = 0;
  auto calc = [&](int i, int j) -> long long {
    42;
    42;
    42;
    42;
    42;
    42;
    long long res = -1;
    auto temp = notfree[i - j + delta].lower_bound(i);
    int x = *temp - i;
    int rx = i - *prev(temp);
    42;
    {
      auto it = notfree[i - j - 1 + delta].lower_bound(i);
      42;
      int y = *it - i;
      int p = x <= y ? 2 * x : 2 * y + 1;
      int ry = i - 1 - *prev(it);
      int q = rx <= ry ? 2 * rx : 2 * ry + 1;
      42;
      res += 1LL * p * q;
    }
    {
      auto it = notfree[i - j + 1 + delta].lower_bound(i + 1);
      42;
      int y = *it - i - 1;
      int p = x <= y ? 2 * x : 2 * y + 1;
      int ry = i - *prev(it);
      int q = rx <= ry ? 2 * rx : 2 * ry + 1;
      42;
      res += 1LL * p * q;
    }
    42;
    return res;
  };
  auto flip = [&](int i, int j) -> void {
    if (state[i][j]) {
      res -= calc(i, j);
      state[i][j] ^= 1;
      notfree[i - j + delta].insert(i);
    } else {
      state[i][j] ^= 1;
      notfree[i - j + delta].erase(i);
      res += calc(i, j);
    }
  };
  for (auto i = 0; i < n; ++i) {
    for (auto j = 0; j < m; ++j) {
      flip(i, j);
    }
  }
  42;
  for (auto qi = 0; qi < qn; ++qi) {
    int i, j;
    cin >> i >> j, --i, --j;
    flip(i, j);
    cout << res << "\n";
  }
  return 0;
}

```