#include <bits/stdc++.h>
using namespace std;
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  long long T, a, b, c, d, n, m, q;
  std::cin >> n >> m >> q;
  std::vector<vector<bool>> state(n, vector<bool>(m, false));
  long long poss = 2 * m - 1;
  std::vector<int> dp(m, 2);
  dp[0] = 1;
  std::vector<int> newDp(m);
  for (int i = 1; i < n; i++) {
    newDp[0] = 1;
    poss += 1;
    for (int j = 1; j < m; j++) {
      newDp[j] = dp[j - 1] + 2;
      poss += dp[j - 1] * 2 + 2;
    }
    poss += dp[m - 1];
    swap(dp, newDp);
  }
  for (int i = 0; i < q; i++) {
    std::cin >> a >> b;
    a--;
    b--;
    long long dir = -1;
    if (state[a][b]) {
      dir = 1;
    }
    state[a][b] = !state[a][b];
    valarray<long long> start[] = {{1, 0}, {-1, 0}, {0, -1}, {0, 1}};
    long long results[] = {0, 0, 0, 0};
    for (int j = 0; j < 4; j++) {
      long long temp[2] = {a, b};
      std::valarray<long long> cur(temp, 2);
      int index = j;
      cur += start[index];
      index = 3 - index;
      while (cur[0] >= 0 && cur[1] >= 0 && cur[0] < n && cur[1] < m &&
             state[cur[0]][cur[1]] == false) {
        results[j]++;
        cur += start[index];
        index = 3 - index;
      }
    }
    poss += results[0] * results[2] * dir;
    poss += results[1] * results[3] * dir;
    poss += results[0] * dir;
    poss += results[1] * dir;
    poss += results[2] * dir;
    poss += results[3] * dir;
    poss += dir;
    cout << poss << endl;
  }
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  long long T, a, b, c, d, n, m, q;
  std::cin >> n >> m >> q;
  std::vector<vector<bool>> state(n, vector<bool>(m, false));
  long long poss = 2 * m - 1;
  std::vector<int> dp(m, 2);
  dp[0] = 1;
  std::vector<int> newDp(m);
  for (int i = 1; i < n; i++) {
    newDp[0] = 1;
    poss += 1;
    for (int j = 1; j < m; j++) {
      newDp[j] = dp[j - 1] + 2;
      poss += dp[j - 1] * 2 + 2;
    }
    poss += dp[m - 1];
    swap(dp, newDp);
  }
  for (int i = 0; i < q; i++) {
    std::cin >> a >> b;
    a--;
    b--;
    long long dir = -1;
    if (state[a][b]) {
      dir = 1;
    }
    state[a][b] = !state[a][b];
    valarray<long long> start[] = {{1, 0}, {-1, 0}, {0, -1}, {0, 1}};
    long long results[] = {0, 0, 0, 0};
    for (int j = 0; j < 4; j++) {
      long long temp[2] = {a, b};
      std::valarray<long long> cur(temp, 2);
      int index = j;
      cur += start[index];
      index = 3 - index;
      while (cur[0] >= 0 && cur[1] >= 0 && cur[0] < n && cur[1] < m &&
             state[cur[0]][cur[1]] == false) {
        results[j]++;
        cur += start[index];
        index = 3 - index;
      }
    }
    poss += results[0] * results[2] * dir;
    poss += results[1] * results[3] * dir;
    poss += results[0] * dir;
    poss += results[1] * dir;
    poss += results[2] * dir;
    poss += results[3] * dir;
    poss += dir;
    cout << poss << endl;
  }
}

```