#include <bits/stdc++.h>
using namespace std;
using LL = long long;
const int MAXN = 1010;
int n, m;
bool z[MAXN][MAXN];
LL ans = 0;
int GetUp(int a, int b, int dir) {
  int ans = 0;
  while (1 <= a && a <= n && 1 <= b && b <= m && !z[a][b]) {
    ++ans;
    if (dir == 1)
      --a;
    else
      --b;
    dir ^= 1;
  }
  return ans;
}
int GetDown(int a, int b, int dir) {
  int ans = 0;
  while (1 <= a && a <= n && 1 <= b && b <= m && !z[a][b]) {
    ++ans;
    if (dir == 1)
      ++a;
    else
      ++b;
    dir ^= 1;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int q;
  cin >> n >> m >> q;
  for (int i = 1; i <= n; ++i) {
    int L = GetDown(i, 1, 1);
    ans += L * (L + 1) / 2;
  }
  for (int i = 1; i <= m; ++i) {
    int L = GetDown(1, i, 0);
    ans += L * (L + 1) / 2;
  }
  ans -= n * m;
  while (q--) {
    int a, b;
    cin >> a >> b;
    if (z[a][b]) {
      int L = GetUp(a, b - 1, 1);
      int R = GetDown(a + 1, b, 0);
      ans -= L * (L + 1) / 2;
      ans -= R * (R + 1) / 2;
      ans += (L + R + 2) * (L + R + 1) / 2;
      L = GetUp(a - 1, b, 0);
      R = GetDown(a, b + 1, 1);
      ans -= L * (L + 1) / 2;
      ans -= R * (R + 1) / 2;
      ans += (L + R + 2) * (L + R + 1) / 2;
      --ans;
    } else {
      int L = GetUp(a, b - 1, 1);
      int R = GetDown(a + 1, b, 0);
      ans += L * (L + 1) / 2;
      ans += R * (R + 1) / 2;
      ans -= (L + R + 2) * (L + R + 1) / 2;
      L = GetUp(a - 1, b, 0);
      R = GetDown(a, b + 1, 1);
      ans += L * (L + 1) / 2;
      ans += R * (R + 1) / 2;
      ans -= (L + R + 2) * (L + R + 1) / 2;
      ++ans;
    }
    z[a][b] = !z[a][b];
    cout << ans << "\n";
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
using LL = long long;
const int MAXN = 1010;
int n, m;
bool z[MAXN][MAXN];
LL ans = 0;
int GetUp(int a, int b, int dir) {
  int ans = 0;
  while (1 <= a && a <= n && 1 <= b && b <= m && !z[a][b]) {
    ++ans;
    if (dir == 1)
      --a;
    else
      --b;
    dir ^= 1;
  }
  return ans;
}
int GetDown(int a, int b, int dir) {
  int ans = 0;
  while (1 <= a && a <= n && 1 <= b && b <= m && !z[a][b]) {
    ++ans;
    if (dir == 1)
      ++a;
    else
      ++b;
    dir ^= 1;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int q;
  cin >> n >> m >> q;
  for (int i = 1; i <= n; ++i) {
    int L = GetDown(i, 1, 1);
    ans += L * (L + 1) / 2;
  }
  for (int i = 1; i <= m; ++i) {
    int L = GetDown(1, i, 0);
    ans += L * (L + 1) / 2;
  }
  ans -= n * m;
  while (q--) {
    int a, b;
    cin >> a >> b;
    if (z[a][b]) {
      int L = GetUp(a, b - 1, 1);
      int R = GetDown(a + 1, b, 0);
      ans -= L * (L + 1) / 2;
      ans -= R * (R + 1) / 2;
      ans += (L + R + 2) * (L + R + 1) / 2;
      L = GetUp(a - 1, b, 0);
      R = GetDown(a, b + 1, 1);
      ans -= L * (L + 1) / 2;
      ans -= R * (R + 1) / 2;
      ans += (L + R + 2) * (L + R + 1) / 2;
      --ans;
    } else {
      int L = GetUp(a, b - 1, 1);
      int R = GetDown(a + 1, b, 0);
      ans += L * (L + 1) / 2;
      ans += R * (R + 1) / 2;
      ans -= (L + R + 2) * (L + R + 1) / 2;
      L = GetUp(a - 1, b, 0);
      R = GetDown(a, b + 1, 1);
      ans += L * (L + 1) / 2;
      ans += R * (R + 1) / 2;
      ans -= (L + R + 2) * (L + R + 1) / 2;
      ++ans;
    }
    z[a][b] = !z[a][b];
    cout << ans << "\n";
  }
  return 0;
}

```