#include <bits/stdc++.h>
using namespace std;
const int DIM = 1000 + 7;
int n, m, q, dp[DIM][DIM], sol;
bool active[DIM][DIM];
int goright[DIM][DIM];
int godown[DIM][DIM];
void upd(int r, int c) {
  if (active[r][c]) {
    sol -= (goright[r][c] + godown[r][c] - 1);
    goright[r][c] = 1 + active[r][c + 1] * (godown[r][c + 1]);
    godown[r][c] = 1 + active[r + 1][c] * (goright[r + 1][c]);
    sol += goright[r][c] + godown[r][c] - 1;
  }
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> q;
  for (int r = n; r >= 1; r--) {
    for (int c = m; c >= 1; c--) {
      goright[r][c] = 1;
      active[r][c] = 1;
      upd(r, c);
    }
  }
  for (int it = 1; it <= q; it++) {
    int r, c;
    cin >> r >> c;
    if (active[r][c]) {
      sol -= (goright[r][c] + godown[r][c] - 1);
    } else {
      goright[r][c] = godown[r][c] = 0;
      goright[r][c] = 1;
    }
    active[r][c] ^= 1;
    vector<pair<int, int>> path;
    path.push_back({r, c});
    int i = r - 1, j = c - 1;
    while (i >= 1 && j >= 1) {
      path.push_back({i, j + 1});
      path.push_back({i + 1, j});
      path.push_back({i, j});
      i--;
      j--;
    }
    if (i >= 1 && j + 1 >= 1) path.push_back({i, j + 1});
    if (i + 1 >= 1 && j >= 1) path.push_back({i + 1, j});
    for (auto &it : path) {
      upd(it.first, it.second);
    }
    cout << sol << "\n";
  }
  return 0;
}

### Prompt

Generate 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;
const int DIM = 1000 + 7;
int n, m, q, dp[DIM][DIM], sol;
bool active[DIM][DIM];
int goright[DIM][DIM];
int godown[DIM][DIM];
void upd(int r, int c) {
  if (active[r][c]) {
    sol -= (goright[r][c] + godown[r][c] - 1);
    goright[r][c] = 1 + active[r][c + 1] * (godown[r][c + 1]);
    godown[r][c] = 1 + active[r + 1][c] * (goright[r + 1][c]);
    sol += goright[r][c] + godown[r][c] - 1;
  }
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> q;
  for (int r = n; r >= 1; r--) {
    for (int c = m; c >= 1; c--) {
      goright[r][c] = 1;
      active[r][c] = 1;
      upd(r, c);
    }
  }
  for (int it = 1; it <= q; it++) {
    int r, c;
    cin >> r >> c;
    if (active[r][c]) {
      sol -= (goright[r][c] + godown[r][c] - 1);
    } else {
      goright[r][c] = godown[r][c] = 0;
      goright[r][c] = 1;
    }
    active[r][c] ^= 1;
    vector<pair<int, int>> path;
    path.push_back({r, c});
    int i = r - 1, j = c - 1;
    while (i >= 1 && j >= 1) {
      path.push_back({i, j + 1});
      path.push_back({i + 1, j});
      path.push_back({i, j});
      i--;
      j--;
    }
    if (i >= 1 && j + 1 >= 1) path.push_back({i, j + 1});
    if (i + 1 >= 1 && j >= 1) path.push_back({i + 1, j});
    for (auto &it : path) {
      upd(it.first, it.second);
    }
    cout << sol << "\n";
  }
  return 0;
}

```