#include <bits/stdc++.h>
using namespace std;
ifstream in("in.in");
ofstream out("out.out");
const int nmax = 1005;
int n, m, q, sol;
int dp[2][nmax][nmax];
vector<vector<int>> a;
void calc(int i, int j) {
  dp[0][i][j] = dp[1][i][j] = 0;
  if (a[i][j] == 0) return;
  if (j < m - 1 && a[i][j + 1]) {
    dp[0][i][j] += dp[1][i][j + 1] + 1;
  }
  if (i < n - 1 && a[i + 1][j]) {
    dp[1][i][j] += dp[0][i + 1][j] + 1;
  }
}
void traverse(int i, int j, int dir) {
  if (i < 0 || j < 0) return;
  int oldDp = dp[0][i][j] + dp[1][i][j];
  calc(i, j);
  int newDp = dp[0][i][j] + dp[1][i][j];
  sol += newDp - oldDp;
  if (dir == 0) {
    traverse(i, j - 1, 1);
  } else {
    traverse(i - 1, j, 0);
  }
}
int main() {
  cin >> n >> m >> q;
  a = vector<vector<int>>(n, vector<int>(m, 1));
  for (int i = n - 1; i >= 0; i--) {
    for (int j = m - 1; j >= 0; j--) {
      calc(i, j);
    }
  }
  sol = 0;
  int nrFree = n * m;
  for (int i = n - 1; i >= 0; i--) {
    for (int j = m - 1; j >= 0; j--) {
      sol += dp[0][i][j] + dp[1][i][j];
    }
  }
  for (int i = 0; i < q; i++) {
    int y, x;
    cin >> y >> x;
    y--;
    x--;
    a[y][x] = 1 - a[y][x];
    if (a[y][x] == 0) {
      sol -= dp[0][y][x] + dp[1][y][x];
      dp[0][y][x] = dp[1][y][x] = 0;
      nrFree--;
    } else {
      calc(y, x);
      sol += dp[0][y][x] + dp[1][y][x];
      nrFree++;
    }
    traverse(y - 1, x, 0);
    traverse(y, x - 1, 1);
    cout << sol + nrFree << "\n";
  }
  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;
ifstream in("in.in");
ofstream out("out.out");
const int nmax = 1005;
int n, m, q, sol;
int dp[2][nmax][nmax];
vector<vector<int>> a;
void calc(int i, int j) {
  dp[0][i][j] = dp[1][i][j] = 0;
  if (a[i][j] == 0) return;
  if (j < m - 1 && a[i][j + 1]) {
    dp[0][i][j] += dp[1][i][j + 1] + 1;
  }
  if (i < n - 1 && a[i + 1][j]) {
    dp[1][i][j] += dp[0][i + 1][j] + 1;
  }
}
void traverse(int i, int j, int dir) {
  if (i < 0 || j < 0) return;
  int oldDp = dp[0][i][j] + dp[1][i][j];
  calc(i, j);
  int newDp = dp[0][i][j] + dp[1][i][j];
  sol += newDp - oldDp;
  if (dir == 0) {
    traverse(i, j - 1, 1);
  } else {
    traverse(i - 1, j, 0);
  }
}
int main() {
  cin >> n >> m >> q;
  a = vector<vector<int>>(n, vector<int>(m, 1));
  for (int i = n - 1; i >= 0; i--) {
    for (int j = m - 1; j >= 0; j--) {
      calc(i, j);
    }
  }
  sol = 0;
  int nrFree = n * m;
  for (int i = n - 1; i >= 0; i--) {
    for (int j = m - 1; j >= 0; j--) {
      sol += dp[0][i][j] + dp[1][i][j];
    }
  }
  for (int i = 0; i < q; i++) {
    int y, x;
    cin >> y >> x;
    y--;
    x--;
    a[y][x] = 1 - a[y][x];
    if (a[y][x] == 0) {
      sol -= dp[0][y][x] + dp[1][y][x];
      dp[0][y][x] = dp[1][y][x] = 0;
      nrFree--;
    } else {
      calc(y, x);
      sol += dp[0][y][x] + dp[1][y][x];
      nrFree++;
    }
    traverse(y - 1, x, 0);
    traverse(y, x - 1, 1);
    cout << sol + nrFree << "\n";
  }
  return 0;
}

```