#include <bits/stdc++.h>
using namespace std;
bool vis[1000 + 1][1000 + 1];
long long n, m, ans;
void Add(long long x, long long y, long long d) {
  long long cnt1 = 1, cnt2 = 1, cnt3 = 1, cnt4 = 1, sx = x, sy = y;
  for (long long op = 0;; op ^= 1) {
    if (!op)
      sx++;
    else
      sy++;
    if (sx > n || sy > m || !vis[sx][sy])
      break;
    else
      cnt1++;
  }
  sx = x, sy = y;
  for (long long op = 1;; op ^= 1) {
    if (!op)
      sx++;
    else
      sy++;
    if (sx > n || sy > m || !vis[sx][sy])
      break;
    else
      cnt2++;
  }
  sx = x, sy = y;
  for (long long op = 0;; op ^= 1) {
    if (!op)
      sx--;
    else
      sy--;
    if (sx < 1 || sy < 1 || !vis[sx][sy])
      break;
    else
      cnt3++;
  }
  sx = x, sy = y;
  for (long long op = 1;; op ^= 1) {
    if (!op)
      sx--;
    else
      sy--;
    if (sx < 1 || sy < 1 || !vis[sx][sy])
      break;
    else
      cnt4++;
  }
  ans += (cnt1 * cnt4 + cnt2 * cnt3 - 1) * d;
  return;
}
signed main() {
  long long q, x, y;
  cin >> n >> m >> q;
  for (long long i = 1; i <= n; ++i)
    for (long long j = 1; j <= m; ++j) vis[i][j] = 1;
  for (long long i = 1; i <= n; ++i)
    for (long long j = 1; j <= m; ++j)
      ans += min(2 * (m - j + 1) - 1, 2 * (n - i + 1)) +
             min(2 * (m - j + 1), 2 * (n - i + 1) - 1) - 1;
  while (q--) {
    cin >> x >> y;
    if (!vis[x][y])
      vis[x][y] = 1, Add(x, y, 1);
    else
      vis[x][y] = 0, Add(x, y, -1);
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
bool vis[1000 + 1][1000 + 1];
long long n, m, ans;
void Add(long long x, long long y, long long d) {
  long long cnt1 = 1, cnt2 = 1, cnt3 = 1, cnt4 = 1, sx = x, sy = y;
  for (long long op = 0;; op ^= 1) {
    if (!op)
      sx++;
    else
      sy++;
    if (sx > n || sy > m || !vis[sx][sy])
      break;
    else
      cnt1++;
  }
  sx = x, sy = y;
  for (long long op = 1;; op ^= 1) {
    if (!op)
      sx++;
    else
      sy++;
    if (sx > n || sy > m || !vis[sx][sy])
      break;
    else
      cnt2++;
  }
  sx = x, sy = y;
  for (long long op = 0;; op ^= 1) {
    if (!op)
      sx--;
    else
      sy--;
    if (sx < 1 || sy < 1 || !vis[sx][sy])
      break;
    else
      cnt3++;
  }
  sx = x, sy = y;
  for (long long op = 1;; op ^= 1) {
    if (!op)
      sx--;
    else
      sy--;
    if (sx < 1 || sy < 1 || !vis[sx][sy])
      break;
    else
      cnt4++;
  }
  ans += (cnt1 * cnt4 + cnt2 * cnt3 - 1) * d;
  return;
}
signed main() {
  long long q, x, y;
  cin >> n >> m >> q;
  for (long long i = 1; i <= n; ++i)
    for (long long j = 1; j <= m; ++j) vis[i][j] = 1;
  for (long long i = 1; i <= n; ++i)
    for (long long j = 1; j <= m; ++j)
      ans += min(2 * (m - j + 1) - 1, 2 * (n - i + 1)) +
             min(2 * (m - j + 1), 2 * (n - i + 1) - 1) - 1;
  while (q--) {
    cin >> x >> y;
    if (!vis[x][y])
      vis[x][y] = 1, Add(x, y, 1);
    else
      vis[x][y] = 0, Add(x, y, -1);
    cout << ans << endl;
  }
  return 0;
}

```