#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long n, m, q;
long long mat[1005][1005] = {};
long long ans = 0;
bool valid(long long x, long long y) {
  if (x < n && y < m && x >= 0 && y >= 0 && mat[x][y] == 0) return true;
  return false;
}
void solve() {
  cin >> n >> m >> q;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < m; j++) {
      long long x = n - i, y = m - j;
      ans += min(2 * x - 1, 2 * y) + min(2 * y - 1, 2 * x) - 1;
    }
  }
  for (long long i = 0; i < q; i++) {
    long long x, y;
    cin >> x >> y;
    x--, y--;
    long long c = 0;
    long long aa = 0, bb = 0, cc = 0, dd = 0;
    long long b = 1;
    bool p = mat[x][y];
    mat[x][y] = 0;
    bool pa = true, push_back = true, pc = true, pd = true;
    for (long long i = x, j = y; i < n && j < m; i++, j++) {
      if (mat[i][j]) break;
      if (pa && valid(i + 1, j))
        aa++;
      else
        pa = false;
      if (pa && valid(i + 1, j) && valid(i + 1, j + 1)) aa++;
      if (push_back && valid(i, j + 1))
        bb++;
      else
        push_back = false;
      if (push_back && valid(i, j + 1) && valid(i + 1, j + 1)) bb++;
    }
    for (long long i = x, j = y; i >= 0 && j >= 0; i--, j--) {
      if (mat[i][j]) break;
      if (pc && valid(i - 1, j)) {
        cc++;
        if (valid(i - 1, j - 1)) cc++;
      } else
        pc = false;
      if (pd && valid(i, j - 1)) {
        dd++;
        if (valid(i - 1, j - 1)) dd++;
      } else
        pd = false;
    }
    c = b + aa + bb + cc + dd + cc * bb + dd * aa;
    if (p) {
      ans += c;
    } else
      ans -= c;
    mat[x][y] = p ^ 1;
    cout << ans << endl;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  t = 1;
  while (t--) solve();
}

### Prompt

Create a solution in Cpp for 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 gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long n, m, q;
long long mat[1005][1005] = {};
long long ans = 0;
bool valid(long long x, long long y) {
  if (x < n && y < m && x >= 0 && y >= 0 && mat[x][y] == 0) return true;
  return false;
}
void solve() {
  cin >> n >> m >> q;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < m; j++) {
      long long x = n - i, y = m - j;
      ans += min(2 * x - 1, 2 * y) + min(2 * y - 1, 2 * x) - 1;
    }
  }
  for (long long i = 0; i < q; i++) {
    long long x, y;
    cin >> x >> y;
    x--, y--;
    long long c = 0;
    long long aa = 0, bb = 0, cc = 0, dd = 0;
    long long b = 1;
    bool p = mat[x][y];
    mat[x][y] = 0;
    bool pa = true, push_back = true, pc = true, pd = true;
    for (long long i = x, j = y; i < n && j < m; i++, j++) {
      if (mat[i][j]) break;
      if (pa && valid(i + 1, j))
        aa++;
      else
        pa = false;
      if (pa && valid(i + 1, j) && valid(i + 1, j + 1)) aa++;
      if (push_back && valid(i, j + 1))
        bb++;
      else
        push_back = false;
      if (push_back && valid(i, j + 1) && valid(i + 1, j + 1)) bb++;
    }
    for (long long i = x, j = y; i >= 0 && j >= 0; i--, j--) {
      if (mat[i][j]) break;
      if (pc && valid(i - 1, j)) {
        cc++;
        if (valid(i - 1, j - 1)) cc++;
      } else
        pc = false;
      if (pd && valid(i, j - 1)) {
        dd++;
        if (valid(i - 1, j - 1)) dd++;
      } else
        pd = false;
    }
    c = b + aa + bb + cc + dd + cc * bb + dd * aa;
    if (p) {
      ans += c;
    } else
      ans -= c;
    mat[x][y] = p ^ 1;
    cout << ans << endl;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  t = 1;
  while (t--) solve();
}

```