#include <bits/stdc++.h>
using namespace std;
template <class T>
void _W(const T &x) {
  cout << x;
}
template <class T>
void _W(T &x) {
  cout << x;
}
template <class T, class U>
void _W(const pair<T, U> &x) {
  _W(x.first);
  putchar(' ');
  _W(x.second);
}
template <class T>
void _W(const vector<T> &x) {
  for (auto i = x.begin(); i != x.end(); _W(*i++)) cout << ' ';
}
void W() {}
template <class T, class... U>
void W(const T &head, const U &...tail) {
  _W(head);
  cout << ", ";
  W(tail...);
}
const int N = 1e3 + 10;
const long long mod = 998244353;
long long n, m;
void solve() {
  int q;
  cin >> n >> m >> q;
  vector<vector<int>> a(n, vector<int>(m, 1));
  long long ans = 0;
  for (int x = 0; x < n; x++)
    for (int y = 0; y < m; y++) {
      if (x == 0) {
        for (int k = 1;; k++) {
          int nx = x + k / 2;
          int ny = y + (k + 1) / 2;
          if (nx == n || ny == m) break;
          ans += k;
        }
      }
      if (y == 0) {
        for (int k = 1;; k++) {
          int nx = x + (k + 1) / 2;
          int ny = y + k / 2;
          if (nx == n || ny == m) break;
          ans += k;
        }
      }
    }
  ans += n * m;
  while (q--) {
    int x, y;
    cin >> x >> y;
    --x, --y;
    for (int i = 0; i < 2; i++) {
      int l = 1, r = 1;
      while (1) {
        int nx = x + (l + i) / 2;
        int ny = y + (l + !i) / 2;
        if (nx == n || ny == m || a[nx][ny] == 0) break;
        ++l;
      }
      while (1) {
        int nx = x - (r + !i) / 2;
        int ny = y - (r + i) / 2;
        if (nx < 0 || ny < 0 || a[nx][ny] == 0) break;
        ++r;
      }
      if (a[x][y] == 0) {
        ans += l * r;
      } else {
        ans -= l * r;
      }
    }
    ans += a[x][y];
    a[x][y] ^= 1;
    ans -= a[x][y];
    cout << ans << "\n";
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t = 1;
  while (t--) solve();
  return 0;
}

### Prompt

Please formulate 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;
template <class T>
void _W(const T &x) {
  cout << x;
}
template <class T>
void _W(T &x) {
  cout << x;
}
template <class T, class U>
void _W(const pair<T, U> &x) {
  _W(x.first);
  putchar(' ');
  _W(x.second);
}
template <class T>
void _W(const vector<T> &x) {
  for (auto i = x.begin(); i != x.end(); _W(*i++)) cout << ' ';
}
void W() {}
template <class T, class... U>
void W(const T &head, const U &...tail) {
  _W(head);
  cout << ", ";
  W(tail...);
}
const int N = 1e3 + 10;
const long long mod = 998244353;
long long n, m;
void solve() {
  int q;
  cin >> n >> m >> q;
  vector<vector<int>> a(n, vector<int>(m, 1));
  long long ans = 0;
  for (int x = 0; x < n; x++)
    for (int y = 0; y < m; y++) {
      if (x == 0) {
        for (int k = 1;; k++) {
          int nx = x + k / 2;
          int ny = y + (k + 1) / 2;
          if (nx == n || ny == m) break;
          ans += k;
        }
      }
      if (y == 0) {
        for (int k = 1;; k++) {
          int nx = x + (k + 1) / 2;
          int ny = y + k / 2;
          if (nx == n || ny == m) break;
          ans += k;
        }
      }
    }
  ans += n * m;
  while (q--) {
    int x, y;
    cin >> x >> y;
    --x, --y;
    for (int i = 0; i < 2; i++) {
      int l = 1, r = 1;
      while (1) {
        int nx = x + (l + i) / 2;
        int ny = y + (l + !i) / 2;
        if (nx == n || ny == m || a[nx][ny] == 0) break;
        ++l;
      }
      while (1) {
        int nx = x - (r + !i) / 2;
        int ny = y - (r + i) / 2;
        if (nx < 0 || ny < 0 || a[nx][ny] == 0) break;
        ++r;
      }
      if (a[x][y] == 0) {
        ans += l * r;
      } else {
        ans -= l * r;
      }
    }
    ans += a[x][y];
    a[x][y] ^= 1;
    ans -= a[x][y];
    cout << ans << "\n";
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t = 1;
  while (t--) solve();
  return 0;
}

```