#include <bits/stdc++.h>
using namespace std;
int n, m, q, d[2][2] = {{-1, 0}, {0, -1}};
bool c[2021][2021];
long long f[2021][2021][2], ans;
long long solve(int x, int y) {
  long long pre[2] = {0, 0}, suf[2] = {0, 0};
  queue<pair<int, int> > q;
  q.push(make_pair(x, y));
  bool t = 0;
  while (!q.empty()) {
    int tx = q.front().first + d[t][0], ty = q.front().second + d[t][1];
    q.pop();
    t ^= 1;
    if (tx > 0 && ty > 0 && !c[tx][ty]) ++pre[0], q.push(make_pair(tx, ty));
  }
  t = 1;
  q.push(make_pair(x, y));
  while (!q.empty()) {
    int tx = q.front().first + d[t][0], ty = q.front().second + d[t][1];
    q.pop();
    t ^= 1;
    if (tx > 0 && ty > 0 && !c[tx][ty]) ++pre[1], q.push(make_pair(tx, ty));
  }
  t = 0;
  q.push(make_pair(x, y));
  while (!q.empty()) {
    int tx = q.front().first - d[t][0], ty = q.front().second - d[t][1];
    q.pop();
    t ^= 1;
    if (tx <= n && ty <= m && !c[tx][ty]) ++suf[0], q.push(make_pair(tx, ty));
  }
  t = 1;
  q.push(make_pair(x, y));
  while (!q.empty()) {
    int tx = q.front().first - d[t][0], ty = q.front().second - d[t][1];
    q.pop();
    t ^= 1;
    if (tx <= n && ty <= m && !c[tx][ty]) ++suf[1], q.push(make_pair(tx, ty));
  }
  return pre[0] * suf[1] + pre[1] * suf[0] + pre[0] + pre[1] + suf[0] + suf[1] +
         1;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> q;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) {
      if (i > 1) f[i][j][1] += f[i - 1][j][0];
      if (j > 1) f[i][j][0] += f[i][j - 1][1];
      ans += f[i][j][1] + f[i][j][0] + 1;
      ++f[i][j][1];
      ++f[i][j][0];
    }
  while (q--) {
    int x, y;
    cin >> x >> y;
    if (c[x][y])
      ans += solve(x, y);
    else
      ans -= solve(x, y);
    c[x][y] ^= 1;
    cout << ans << '\n';
  }
  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;
int n, m, q, d[2][2] = {{-1, 0}, {0, -1}};
bool c[2021][2021];
long long f[2021][2021][2], ans;
long long solve(int x, int y) {
  long long pre[2] = {0, 0}, suf[2] = {0, 0};
  queue<pair<int, int> > q;
  q.push(make_pair(x, y));
  bool t = 0;
  while (!q.empty()) {
    int tx = q.front().first + d[t][0], ty = q.front().second + d[t][1];
    q.pop();
    t ^= 1;
    if (tx > 0 && ty > 0 && !c[tx][ty]) ++pre[0], q.push(make_pair(tx, ty));
  }
  t = 1;
  q.push(make_pair(x, y));
  while (!q.empty()) {
    int tx = q.front().first + d[t][0], ty = q.front().second + d[t][1];
    q.pop();
    t ^= 1;
    if (tx > 0 && ty > 0 && !c[tx][ty]) ++pre[1], q.push(make_pair(tx, ty));
  }
  t = 0;
  q.push(make_pair(x, y));
  while (!q.empty()) {
    int tx = q.front().first - d[t][0], ty = q.front().second - d[t][1];
    q.pop();
    t ^= 1;
    if (tx <= n && ty <= m && !c[tx][ty]) ++suf[0], q.push(make_pair(tx, ty));
  }
  t = 1;
  q.push(make_pair(x, y));
  while (!q.empty()) {
    int tx = q.front().first - d[t][0], ty = q.front().second - d[t][1];
    q.pop();
    t ^= 1;
    if (tx <= n && ty <= m && !c[tx][ty]) ++suf[1], q.push(make_pair(tx, ty));
  }
  return pre[0] * suf[1] + pre[1] * suf[0] + pre[0] + pre[1] + suf[0] + suf[1] +
         1;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> q;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) {
      if (i > 1) f[i][j][1] += f[i - 1][j][0];
      if (j > 1) f[i][j][0] += f[i][j - 1][1];
      ans += f[i][j][1] + f[i][j][0] + 1;
      ++f[i][j][1];
      ++f[i][j][0];
    }
  while (q--) {
    int x, y;
    cin >> x >> y;
    if (c[x][y])
      ans += solve(x, y);
    else
      ans -= solve(x, y);
    c[x][y] ^= 1;
    cout << ans << '\n';
  }
  return 0;
}

```