#include <bits/stdc++.h>
using namespace std;
mt19937_64 rnd(time(NULL));
const long long mod = 1000000007;
const int N = 1010;
int a[N][N];
int cnt[N][N][2];
int ans = 0;
int n, m, q;
void calc(int i, int j) {
  if (i <= 0 || j <= 0 || i > n || j > m) return;
  ans -= cnt[i][j][0] + cnt[i][j][1];
  if (a[i][j])
    cnt[i][j][0] = cnt[i][j][1] = 0;
  else {
    cnt[i][j][0] = cnt[i - 1][j][1] + 1;
    cnt[i][j][1] = cnt[i][j - 1][0] + 1;
  }
  ans += cnt[i][j][0] + cnt[i][j][1];
}
void solve() {
  cin >> n >> m >> q;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      calc(i, j);
      ans--;
    }
  }
  for (int it = 1; it <= q; it++) {
    int x, y;
    cin >> x >> y;
    ans += 1 - a[x][y];
    a[x][y] ^= 1;
    ans -= 1 - a[x][y];
    for (int i = 0; i <= min(n - x, m - y); i++) {
      calc(x + i, y + i);
      calc(x + i + 1, y + i);
      calc(x + i, y + i + 1);
    }
    cout << ans << '\n';
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int tt = 1;
  while (tt--) {
    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;
mt19937_64 rnd(time(NULL));
const long long mod = 1000000007;
const int N = 1010;
int a[N][N];
int cnt[N][N][2];
int ans = 0;
int n, m, q;
void calc(int i, int j) {
  if (i <= 0 || j <= 0 || i > n || j > m) return;
  ans -= cnt[i][j][0] + cnt[i][j][1];
  if (a[i][j])
    cnt[i][j][0] = cnt[i][j][1] = 0;
  else {
    cnt[i][j][0] = cnt[i - 1][j][1] + 1;
    cnt[i][j][1] = cnt[i][j - 1][0] + 1;
  }
  ans += cnt[i][j][0] + cnt[i][j][1];
}
void solve() {
  cin >> n >> m >> q;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      calc(i, j);
      ans--;
    }
  }
  for (int it = 1; it <= q; it++) {
    int x, y;
    cin >> x >> y;
    ans += 1 - a[x][y];
    a[x][y] ^= 1;
    ans -= 1 - a[x][y];
    for (int i = 0; i <= min(n - x, m - y); i++) {
      calc(x + i, y + i);
      calc(x + i + 1, y + i);
      calc(x + i, y + i + 1);
    }
    cout << ans << '\n';
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int tt = 1;
  while (tt--) {
    solve();
  }
  return 0;
}

```