#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
long long int gcd(long long int a, long long int b) {
  if (!b)
    return a;
  else
    return gcd(b, a % b);
}
long long int dp[3][1001][1001];
bool has[1000][1000];
int n, m, q;
void tran(int i, int j) {
  if (i >= n || i < 0 || j < 0 || j >= m || has[i][j]) return;
  dp[2][i][j] = 1;
  if (i) dp[1][i][j] += dp[2][i - 1][j] + dp[0][i - 1][j];
  if (j) dp[0][i][j] += dp[2][i][j - 1] + dp[1][i][j - 1];
}
long long int ans = 0;
void upd(int i, int j, bool mode) {
  if (i >= n || i < 0 || j < 0 || j >= m) return;
  long long int c = mode ? 1 : -1;
  ans += c * (dp[0][i][j] + dp[1][i][j] + dp[2][i][j]);
  if (!mode) dp[0][i][j] = dp[1][i][j] = dp[2][i][j] = 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> q;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      tran(i, j);
      upd(i, j, 1);
    }
  }
  while (q--) {
    int i, j;
    cin >> i >> j;
    has[i - 1][j - 1] ^= 1;
    while (i && j) {
      i--;
      j--;
    }
    for (int k = 0; k < max(n, m); k++) {
      upd(i + k - 1, j + k, 0);
      upd(i + k, j + k, 0);
      upd(i + k + 1, j + k, 0);
    }
    for (int k = 0; k < max(n, m); k++) {
      tran(i + k - 1, j + k);
      tran(i + k, j + k);
      tran(i + k + 1, j + k);
      upd(i + k - 1, j + k, 1);
      upd(i + k, j + k, 1);
      upd(i + k + 1, j + k, 1);
    }
    cout << ans << '\n';
  }
}

### Prompt

Your task is to create 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>
#pragma GCC optimize("Ofast")
using namespace std;
long long int gcd(long long int a, long long int b) {
  if (!b)
    return a;
  else
    return gcd(b, a % b);
}
long long int dp[3][1001][1001];
bool has[1000][1000];
int n, m, q;
void tran(int i, int j) {
  if (i >= n || i < 0 || j < 0 || j >= m || has[i][j]) return;
  dp[2][i][j] = 1;
  if (i) dp[1][i][j] += dp[2][i - 1][j] + dp[0][i - 1][j];
  if (j) dp[0][i][j] += dp[2][i][j - 1] + dp[1][i][j - 1];
}
long long int ans = 0;
void upd(int i, int j, bool mode) {
  if (i >= n || i < 0 || j < 0 || j >= m) return;
  long long int c = mode ? 1 : -1;
  ans += c * (dp[0][i][j] + dp[1][i][j] + dp[2][i][j]);
  if (!mode) dp[0][i][j] = dp[1][i][j] = dp[2][i][j] = 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> q;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      tran(i, j);
      upd(i, j, 1);
    }
  }
  while (q--) {
    int i, j;
    cin >> i >> j;
    has[i - 1][j - 1] ^= 1;
    while (i && j) {
      i--;
      j--;
    }
    for (int k = 0; k < max(n, m); k++) {
      upd(i + k - 1, j + k, 0);
      upd(i + k, j + k, 0);
      upd(i + k + 1, j + k, 0);
    }
    for (int k = 0; k < max(n, m); k++) {
      tran(i + k - 1, j + k);
      tran(i + k, j + k);
      tran(i + k + 1, j + k);
      upd(i + k - 1, j + k, 1);
      upd(i + k, j + k, 1);
      upd(i + k + 1, j + k, 1);
    }
    cout << ans << '\n';
  }
}

```