#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int M = 200005;
string s, p;
long long upCount[1005][1005], leftCount[1005][1005], a[1005][1005], n, m, q;
long long dfs(int x, int y, int f) {
  if (x == 0 || y == 0 || a[x][y]) return 0;
  if (f) {
    return 1 + dfs(x - 1, y, f ^ 1);
  } else {
    return 1 + dfs(x, y - 1, f ^ 1);
  }
}
long long dfs2(int x, int y, int f) {
  if (x > n || y > m || a[x][y]) return 0;
  if (f) {
    return 1 + dfs2(x + 1, y, f ^ 1);
  } else {
    return 1 + dfs2(x, y + 1, f ^ 1);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> q;
  long long res = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      leftCount[i][j] = upCount[i][j - 1] + 1;
      upCount[i][j] = leftCount[i - 1][j] + 1;
      res += leftCount[i][j];
      res += upCount[i][j];
      res -= 1;
    }
  }
  while (q--) {
    int x, y;
    cin >> x >> y;
    long long up = dfs(x - 1, y, 0);
    long long down = dfs2(x + 1, y, 0);
    long long lft = dfs(x, y - 1, 1);
    long long rgt = dfs2(x, y + 1, 1);
    if (a[x][y]) {
      res += (up + 1) * (rgt + 1);
      res += (lft + 1) * (down + 1);
      res -= 1;
    } else {
      res -= (up + 1) * (rgt + 1);
      res -= (lft + 1) * (down + 1);
      res += 1;
    }
    a[x][y] ^= 1;
    cout << res << "\n";
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 rng(chrono::steady_clock::now().time_since_epoch().count());
const int M = 200005;
string s, p;
long long upCount[1005][1005], leftCount[1005][1005], a[1005][1005], n, m, q;
long long dfs(int x, int y, int f) {
  if (x == 0 || y == 0 || a[x][y]) return 0;
  if (f) {
    return 1 + dfs(x - 1, y, f ^ 1);
  } else {
    return 1 + dfs(x, y - 1, f ^ 1);
  }
}
long long dfs2(int x, int y, int f) {
  if (x > n || y > m || a[x][y]) return 0;
  if (f) {
    return 1 + dfs2(x + 1, y, f ^ 1);
  } else {
    return 1 + dfs2(x, y + 1, f ^ 1);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> q;
  long long res = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      leftCount[i][j] = upCount[i][j - 1] + 1;
      upCount[i][j] = leftCount[i - 1][j] + 1;
      res += leftCount[i][j];
      res += upCount[i][j];
      res -= 1;
    }
  }
  while (q--) {
    int x, y;
    cin >> x >> y;
    long long up = dfs(x - 1, y, 0);
    long long down = dfs2(x + 1, y, 0);
    long long lft = dfs(x, y - 1, 1);
    long long rgt = dfs2(x, y + 1, 1);
    if (a[x][y]) {
      res += (up + 1) * (rgt + 1);
      res += (lft + 1) * (down + 1);
      res -= 1;
    } else {
      res -= (up + 1) * (rgt + 1);
      res -= (lft + 1) * (down + 1);
      res += 1;
    }
    a[x][y] ^= 1;
    cout << res << "\n";
  }
  return 0;
}

```