#include <bits/stdc++.h>
using namespace std;
long long cnt[1005][1005][2];
bool vis[1005][1005];
long long n, m, ans;
long long getadd(long long x, long long y, int dir) {
  if (x >= n || y >= m) return 0;
  if (vis[x][y]) return 0;
  if (dir == 0) return getadd(x, y + 1, 1) + 1;
  return getadd(x + 1, y, 0) + 1;
}
void dfs(long long x, long long y, long long k, int dir) {
  if (x < 0 || y < 0) return;
  if (vis[x][y]) return;
  if (dir == 0) {
    cnt[x][y][0] += k;
    ans += k;
    dfs(x - 1, y, k, 1);
  } else {
    cnt[x][y][1] += k;
    ans += k;
    dfs(x, y - 1, k, 0);
  }
}
int main() {
  long long i, j, q, x, y, k0, k1;
  scanf("%lld%lld%lld", &n, &m, &q);
  ans = 0;
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      cnt[i][j][0] = min(n - i, m - j) * 2 - 1;
      cnt[i][j][1] = min(n - i, m - j) * 2 - 1;
      if (n - i > m - j) cnt[i][j][1]++;
      if (n - i < m - j) cnt[i][j][0]++;
      ans += cnt[i][j][0] + cnt[i][j][1] - 1;
    }
  }
  while (q--) {
    scanf("%lld%lld", &x, &y);
    if (vis[x - 1][y - 1]) {
      vis[x - 1][y - 1] = false;
      k1 = getadd(x - 1, y - 1, 0);
      k0 = getadd(x - 1, y - 1, 1);
      dfs(x - 1, y - 1, k1, 0);
      dfs(x - 1, y - 1, k0, 1);
      ans--;
    } else {
      k0 = cnt[x - 1][y - 1][0];
      k1 = cnt[x - 1][y - 1][1];
      cnt[x - 1][y - 1][0] = 0;
      cnt[x - 1][y - 1][1] = 0;
      ans -= k0 + k1 - 1;
      dfs(x - 1, y - 2, -k1, 0);
      dfs(x - 2, y - 1, -k0, 1);
      vis[x - 1][y - 1] = true;
    }
    printf("%lld\n", ans);
  }
  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;
long long cnt[1005][1005][2];
bool vis[1005][1005];
long long n, m, ans;
long long getadd(long long x, long long y, int dir) {
  if (x >= n || y >= m) return 0;
  if (vis[x][y]) return 0;
  if (dir == 0) return getadd(x, y + 1, 1) + 1;
  return getadd(x + 1, y, 0) + 1;
}
void dfs(long long x, long long y, long long k, int dir) {
  if (x < 0 || y < 0) return;
  if (vis[x][y]) return;
  if (dir == 0) {
    cnt[x][y][0] += k;
    ans += k;
    dfs(x - 1, y, k, 1);
  } else {
    cnt[x][y][1] += k;
    ans += k;
    dfs(x, y - 1, k, 0);
  }
}
int main() {
  long long i, j, q, x, y, k0, k1;
  scanf("%lld%lld%lld", &n, &m, &q);
  ans = 0;
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      cnt[i][j][0] = min(n - i, m - j) * 2 - 1;
      cnt[i][j][1] = min(n - i, m - j) * 2 - 1;
      if (n - i > m - j) cnt[i][j][1]++;
      if (n - i < m - j) cnt[i][j][0]++;
      ans += cnt[i][j][0] + cnt[i][j][1] - 1;
    }
  }
  while (q--) {
    scanf("%lld%lld", &x, &y);
    if (vis[x - 1][y - 1]) {
      vis[x - 1][y - 1] = false;
      k1 = getadd(x - 1, y - 1, 0);
      k0 = getadd(x - 1, y - 1, 1);
      dfs(x - 1, y - 1, k1, 0);
      dfs(x - 1, y - 1, k0, 1);
      ans--;
    } else {
      k0 = cnt[x - 1][y - 1][0];
      k1 = cnt[x - 1][y - 1][1];
      cnt[x - 1][y - 1][0] = 0;
      cnt[x - 1][y - 1][1] = 0;
      ans -= k0 + k1 - 1;
      dfs(x - 1, y - 2, -k1, 0);
      dfs(x - 2, y - 1, -k0, 1);
      vis[x - 1][y - 1] = true;
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```