#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m, q;
  cin >> n >> m >> q;
  vector<vector<int>> a(n, vector<int>(m, 1));
  vector<vector<int>> b(n, vector<int>(m));
  for (int i = n - 1; i >= 0; i--) {
    for (int j = m - 1; j >= 0; j--) {
      if (a[i][j]) {
        b[i][j] = max(b[i][j], i + j);
        if (i && j && a[i - 1][j - 1]) {
          b[i - 1][j - 1] = b[i][j];
        }
      }
    }
  }
  long long ans = n * m;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (!a[i][j]) {
        continue;
      }
      if (i) {
        if (a[i - 1][j]) {
          int A = (b[i][j] - i - j) / 2 + 1;
          int B = (b[i - 1][j] - i + 1 - j) / 2 + 1;
          if (A == B) {
            ans += A + B - 1;
          } else if (A < B) {
            ans += A * 2;
          } else {
            ans += B * 2 - 1;
          }
        }
      }
      if (j + 1 < m) {
        if (a[i][j + 1]) {
          int A = (b[i][j] - i - j) / 2 + 1;
          int B = (b[i][j + 1] - i - j - 1) / 2 + 1;
          if (A == B) {
            ans += A + B - 1;
          } else if (A > B) {
            ans += B * 2;
          } else {
            ans += A * 2 - 1;
          }
        }
      }
    }
  }
  for (int i = 0; i < q; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    int oldx = x, oldy = y;
    int p = min(n - x - 1, m - y - 1);
    x += p;
    y += p;
    vector<pair<int, int>> tmp;
    while (x >= 0 && y >= 0) {
      if (x + 1 < n && y == m - 1) {
        tmp.push_back({x + 1, y});
      }
      if (y - 1 >= 0) {
        tmp.push_back({x, y - 1});
      }
      tmp.push_back({x, y});
      x--;
      y--;
    }
    for (auto j : tmp) {
      b[j.first][j.second] = 0;
    }
    for (auto j : tmp) {
      if (a[j.first][j.second]) {
        b[j.first][j.second] = max(b[j.first][j.second], j.first + j.second);
        if (j.first && j.second && a[j.first - 1][j.second - 1]) {
          b[j.first - 1][j.second - 1] = b[j.first][j.second];
        }
      }
    }
    for (auto j : tmp) {
      if (!a[j.first][j.second]) {
        continue;
      }
      if (j.first) {
        if (a[j.first - 1][j.second]) {
          int A = (b[j.first][j.second] - j.first - j.second) / 2 + 1;
          int B = (b[j.first - 1][j.second] - j.first + 1 - j.second) / 2 + 1;
          if (A == B) {
            ans -= A + B - 1;
          } else if (A < B) {
            ans -= A * 2;
          } else {
            ans -= B * 2 - 1;
          }
        }
      }
      if (j.second + 1 < m) {
        if (a[j.first][j.second + 1]) {
          int A = (b[j.first][j.second] - j.first - j.second) / 2 + 1;
          int B = (b[j.first][j.second + 1] - j.first - j.second - 1) / 2 + 1;
          if (A == B) {
            ans -= A + B - 1;
          } else if (A > B) {
            ans -= B * 2;
          } else {
            ans -= A * 2 - 1;
          }
        }
      }
    }
    if (a[oldx][oldy] == 1) {
      ans--;
    } else {
      ans++;
    }
    a[oldx][oldy] ^= 1;
    for (auto j : tmp) {
      b[j.first][j.second] = 0;
    }
    for (auto j : tmp) {
      if (a[j.first][j.second]) {
        b[j.first][j.second] = max(b[j.first][j.second], j.first + j.second);
        if (j.first && j.second && a[j.first - 1][j.second - 1]) {
          b[j.first - 1][j.second - 1] = b[j.first][j.second];
        }
      }
    }
    for (auto j : tmp) {
      if (!a[j.first][j.second]) {
        continue;
      }
      if (j.first) {
        if (a[j.first - 1][j.second]) {
          int A = (b[j.first][j.second] - j.first - j.second) / 2 + 1;
          int B = (b[j.first - 1][j.second] - j.first + 1 - j.second) / 2 + 1;
          if (A == B) {
            ans += A + B - 1;
          } else if (A < B) {
            ans += A * 2;
          } else {
            ans += B * 2 - 1;
          }
        }
      }
      if (j.second + 1 < m) {
        if (a[j.first][j.second + 1]) {
          int A = (b[j.first][j.second] - j.first - j.second) / 2 + 1;
          int B = (b[j.first][j.second + 1] - j.first - j.second - 1) / 2 + 1;
          if (A == B) {
            ans += A + B - 1;
          } else if (A > B) {
            ans += B * 2;
          } else {
            ans += A * 2 - 1;
          }
        }
      }
    }
    cout << ans << endl;
  }
}

### 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>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m, q;
  cin >> n >> m >> q;
  vector<vector<int>> a(n, vector<int>(m, 1));
  vector<vector<int>> b(n, vector<int>(m));
  for (int i = n - 1; i >= 0; i--) {
    for (int j = m - 1; j >= 0; j--) {
      if (a[i][j]) {
        b[i][j] = max(b[i][j], i + j);
        if (i && j && a[i - 1][j - 1]) {
          b[i - 1][j - 1] = b[i][j];
        }
      }
    }
  }
  long long ans = n * m;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (!a[i][j]) {
        continue;
      }
      if (i) {
        if (a[i - 1][j]) {
          int A = (b[i][j] - i - j) / 2 + 1;
          int B = (b[i - 1][j] - i + 1 - j) / 2 + 1;
          if (A == B) {
            ans += A + B - 1;
          } else if (A < B) {
            ans += A * 2;
          } else {
            ans += B * 2 - 1;
          }
        }
      }
      if (j + 1 < m) {
        if (a[i][j + 1]) {
          int A = (b[i][j] - i - j) / 2 + 1;
          int B = (b[i][j + 1] - i - j - 1) / 2 + 1;
          if (A == B) {
            ans += A + B - 1;
          } else if (A > B) {
            ans += B * 2;
          } else {
            ans += A * 2 - 1;
          }
        }
      }
    }
  }
  for (int i = 0; i < q; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    int oldx = x, oldy = y;
    int p = min(n - x - 1, m - y - 1);
    x += p;
    y += p;
    vector<pair<int, int>> tmp;
    while (x >= 0 && y >= 0) {
      if (x + 1 < n && y == m - 1) {
        tmp.push_back({x + 1, y});
      }
      if (y - 1 >= 0) {
        tmp.push_back({x, y - 1});
      }
      tmp.push_back({x, y});
      x--;
      y--;
    }
    for (auto j : tmp) {
      b[j.first][j.second] = 0;
    }
    for (auto j : tmp) {
      if (a[j.first][j.second]) {
        b[j.first][j.second] = max(b[j.first][j.second], j.first + j.second);
        if (j.first && j.second && a[j.first - 1][j.second - 1]) {
          b[j.first - 1][j.second - 1] = b[j.first][j.second];
        }
      }
    }
    for (auto j : tmp) {
      if (!a[j.first][j.second]) {
        continue;
      }
      if (j.first) {
        if (a[j.first - 1][j.second]) {
          int A = (b[j.first][j.second] - j.first - j.second) / 2 + 1;
          int B = (b[j.first - 1][j.second] - j.first + 1 - j.second) / 2 + 1;
          if (A == B) {
            ans -= A + B - 1;
          } else if (A < B) {
            ans -= A * 2;
          } else {
            ans -= B * 2 - 1;
          }
        }
      }
      if (j.second + 1 < m) {
        if (a[j.first][j.second + 1]) {
          int A = (b[j.first][j.second] - j.first - j.second) / 2 + 1;
          int B = (b[j.first][j.second + 1] - j.first - j.second - 1) / 2 + 1;
          if (A == B) {
            ans -= A + B - 1;
          } else if (A > B) {
            ans -= B * 2;
          } else {
            ans -= A * 2 - 1;
          }
        }
      }
    }
    if (a[oldx][oldy] == 1) {
      ans--;
    } else {
      ans++;
    }
    a[oldx][oldy] ^= 1;
    for (auto j : tmp) {
      b[j.first][j.second] = 0;
    }
    for (auto j : tmp) {
      if (a[j.first][j.second]) {
        b[j.first][j.second] = max(b[j.first][j.second], j.first + j.second);
        if (j.first && j.second && a[j.first - 1][j.second - 1]) {
          b[j.first - 1][j.second - 1] = b[j.first][j.second];
        }
      }
    }
    for (auto j : tmp) {
      if (!a[j.first][j.second]) {
        continue;
      }
      if (j.first) {
        if (a[j.first - 1][j.second]) {
          int A = (b[j.first][j.second] - j.first - j.second) / 2 + 1;
          int B = (b[j.first - 1][j.second] - j.first + 1 - j.second) / 2 + 1;
          if (A == B) {
            ans += A + B - 1;
          } else if (A < B) {
            ans += A * 2;
          } else {
            ans += B * 2 - 1;
          }
        }
      }
      if (j.second + 1 < m) {
        if (a[j.first][j.second + 1]) {
          int A = (b[j.first][j.second] - j.first - j.second) / 2 + 1;
          int B = (b[j.first][j.second + 1] - j.first - j.second - 1) / 2 + 1;
          if (A == B) {
            ans += A + B - 1;
          } else if (A > B) {
            ans += B * 2;
          } else {
            ans += A * 2 - 1;
          }
        }
      }
    }
    cout << ans << endl;
  }
}

```