#include <bits/stdc++.h>
using namespace std;
const long long N = 1e3 + 10;
long long dp1[N][N], dp2[N][N];
long long a[N][N];
signed main() {
  ios::sync_with_stdio(false);
  long long n, m, q;
  cin >> n >> m >> q;
  for (long long i = n; i >= 1; i--)
    for (long long j = m; j >= 1; j--) {
      dp1[i][j] = dp2[i][j + 1] + 1;
      dp2[i][j] = dp1[i + 1][j] + 1;
    }
  long long ans = 0;
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= m; j++) ans += dp1[i][j] + dp2[i][j];
  long long cnt = n * m;
  while (q--) {
    long long x, y;
    cin >> x >> y;
    bool flag = false;
    if (a[x][y] == 0) {
      cnt--;
      flag = 1;
    } else {
      cnt++;
    }
    a[x][y] ^= 1;
    vector<pair<long long, long long> > tmp;
    tmp.push_back(make_pair(x, y));
    long long i = x, j = y + 1;
    while (true) {
      if (i > n || j > m) break;
      tmp.push_back(make_pair(i, j));
      i = i + 1;
      if (i > n || j > m) break;
      tmp.push_back(make_pair(i, j));
      j = j + 1;
    }
    reverse(tmp.begin(), tmp.end());
    i = x - 1, j = y;
    while (true) {
      if (i <= 0 || j <= 0) break;
      tmp.push_back(make_pair(i, j));
      j = j - 1;
      if (i <= 0 || j <= 0) break;
      tmp.push_back(make_pair(i, j));
      i = i - 1;
    }
    for (pair<long long, long long> v : tmp) {
      long long i = v.first, j = v.second;
      if (a[i][j] == 1) {
        ans -= dp1[i][j] + dp2[i][j];
        dp1[i][j] = 0, dp2[i][j] = 0;
      } else {
        ans -= dp1[i][j] + dp2[i][j];
        dp1[i][j] = dp2[i][j + 1] + 1;
        dp2[i][j] = dp1[i + 1][j] + 1;
        ans += dp1[i][j] + dp2[i][j];
      }
    }
    tmp.clear();
    tmp.push_back(make_pair(x, y));
    i = x + 1, j = y;
    while (true) {
      if (i > n || j > m) break;
      tmp.push_back(make_pair(i, j));
      j = j + 1;
      if (i > n || j > m) break;
      tmp.push_back(make_pair(i, j));
      i = i + 1;
    }
    reverse(tmp.begin(), tmp.end());
    i = x, j = y - 1;
    while (true) {
      if (i <= 0 || j <= 0) break;
      tmp.push_back(make_pair(i, j));
      i = i - 1;
      if (i <= 0 || j <= 0) break;
      tmp.push_back(make_pair(i, j));
      j = j - 1;
    }
    for (pair<long long, long long> v : tmp) {
      long long i = v.first, j = v.second;
      if (a[i][j] == 1) {
        ans -= dp1[i][j] + dp2[i][j];
        dp1[i][j] = 0, dp2[i][j] = 0;
      } else {
        ans -= dp1[i][j] + dp2[i][j];
        dp1[i][j] = dp2[i][j + 1] + 1;
        dp2[i][j] = dp1[i + 1][j] + 1;
        ans += dp1[i][j] + dp2[i][j];
      }
    }
    cout << ans - cnt << endl;
  }
  return 0;
}

### 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;
const long long N = 1e3 + 10;
long long dp1[N][N], dp2[N][N];
long long a[N][N];
signed main() {
  ios::sync_with_stdio(false);
  long long n, m, q;
  cin >> n >> m >> q;
  for (long long i = n; i >= 1; i--)
    for (long long j = m; j >= 1; j--) {
      dp1[i][j] = dp2[i][j + 1] + 1;
      dp2[i][j] = dp1[i + 1][j] + 1;
    }
  long long ans = 0;
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= m; j++) ans += dp1[i][j] + dp2[i][j];
  long long cnt = n * m;
  while (q--) {
    long long x, y;
    cin >> x >> y;
    bool flag = false;
    if (a[x][y] == 0) {
      cnt--;
      flag = 1;
    } else {
      cnt++;
    }
    a[x][y] ^= 1;
    vector<pair<long long, long long> > tmp;
    tmp.push_back(make_pair(x, y));
    long long i = x, j = y + 1;
    while (true) {
      if (i > n || j > m) break;
      tmp.push_back(make_pair(i, j));
      i = i + 1;
      if (i > n || j > m) break;
      tmp.push_back(make_pair(i, j));
      j = j + 1;
    }
    reverse(tmp.begin(), tmp.end());
    i = x - 1, j = y;
    while (true) {
      if (i <= 0 || j <= 0) break;
      tmp.push_back(make_pair(i, j));
      j = j - 1;
      if (i <= 0 || j <= 0) break;
      tmp.push_back(make_pair(i, j));
      i = i - 1;
    }
    for (pair<long long, long long> v : tmp) {
      long long i = v.first, j = v.second;
      if (a[i][j] == 1) {
        ans -= dp1[i][j] + dp2[i][j];
        dp1[i][j] = 0, dp2[i][j] = 0;
      } else {
        ans -= dp1[i][j] + dp2[i][j];
        dp1[i][j] = dp2[i][j + 1] + 1;
        dp2[i][j] = dp1[i + 1][j] + 1;
        ans += dp1[i][j] + dp2[i][j];
      }
    }
    tmp.clear();
    tmp.push_back(make_pair(x, y));
    i = x + 1, j = y;
    while (true) {
      if (i > n || j > m) break;
      tmp.push_back(make_pair(i, j));
      j = j + 1;
      if (i > n || j > m) break;
      tmp.push_back(make_pair(i, j));
      i = i + 1;
    }
    reverse(tmp.begin(), tmp.end());
    i = x, j = y - 1;
    while (true) {
      if (i <= 0 || j <= 0) break;
      tmp.push_back(make_pair(i, j));
      i = i - 1;
      if (i <= 0 || j <= 0) break;
      tmp.push_back(make_pair(i, j));
      j = j - 1;
    }
    for (pair<long long, long long> v : tmp) {
      long long i = v.first, j = v.second;
      if (a[i][j] == 1) {
        ans -= dp1[i][j] + dp2[i][j];
        dp1[i][j] = 0, dp2[i][j] = 0;
      } else {
        ans -= dp1[i][j] + dp2[i][j];
        dp1[i][j] = dp2[i][j + 1] + 1;
        dp2[i][j] = dp1[i + 1][j] + 1;
        ans += dp1[i][j] + dp2[i][j];
      }
    }
    cout << ans - cnt << endl;
  }
  return 0;
}

```