#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
const int ms = 1000 + 5;
const int mod = 1e9 + 7;
using ii = pair<int, int>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int a[ms][ms];
int n, m, q;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
int get(int x, int y, int d1, int d2) {
  if (x < 0 || x >= n || y < 0 || y >= m || a[x][y]) {
    return 0;
  }
  return 1 + get(x + dx[d1], y + dy[d1], d2, d1);
}
void solve() {
  cin >> n >> m >> q;
  int ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      int dx = n - i - 1;
      int dy = m - j - 1;
      ans++;
      ans += min(dx, dy) * 4;
      if (dx != dy) ans++;
    }
  }
  while (q--) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    int tmp = a[x][y];
    a[x][y] = 0;
    int cur = get(x, y, 1, 0) * get(x, y, 2, 3) +
              get(x, y, 0, 1) * get(x, y, 3, 2) - 1;
    if (!tmp) {
      ans -= cur;
    } else {
      ans += cur;
    }
    a[x][y] = !tmp;
    cout << ans << '\n';
  }
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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;
const int inf = 1e9;
const int ms = 1000 + 5;
const int mod = 1e9 + 7;
using ii = pair<int, int>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int a[ms][ms];
int n, m, q;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
int get(int x, int y, int d1, int d2) {
  if (x < 0 || x >= n || y < 0 || y >= m || a[x][y]) {
    return 0;
  }
  return 1 + get(x + dx[d1], y + dy[d1], d2, d1);
}
void solve() {
  cin >> n >> m >> q;
  int ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      int dx = n - i - 1;
      int dy = m - j - 1;
      ans++;
      ans += min(dx, dy) * 4;
      if (dx != dy) ans++;
    }
  }
  while (q--) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    int tmp = a[x][y];
    a[x][y] = 0;
    int cur = get(x, y, 1, 0) * get(x, y, 2, 3) +
              get(x, y, 0, 1) * get(x, y, 3, 2) - 1;
    if (!tmp) {
      ans -= cur;
    } else {
      ans += cur;
    }
    a[x][y] = !tmp;
    cout << ans << '\n';
  }
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```