#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int dx[] = {0, 0, 1, -1}, dy[] = {1, -1, 0, 0};
void debug(int x = 0) { cout << "--ok(" << x << ")" << endl; }
const int N = 1e3 + 5;
long long f[N][N][2];
int g[N][N];
long long ans = 0, tot;
void work(int l, int r, int k) {
  if (l < 1 || r < 1) return;
  if (g[l][r] == 0) return;
  long long now = f[l][r][k];
  if (k == 0) {
    f[l][r][k] = f[l][r + 1][1] + 1;
    ans = ans + f[l][r][k] - now;
    work(l - 1, r, 1);
  } else {
    f[l][r][k] = f[l + 1][r][0] + 1;
    ans = ans + f[l][r][k] - now;
    work(l, r - 1, 0);
  }
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios::sync_with_stdio(false);
  int n, m, q;
  cin >> n >> m >> q;
  for (int i = n; i >= 1; i--) {
    for (int j = m; j >= 1; j--) {
      f[i][j][0] = 1 + f[i][j + 1][1];
      f[i][j][1] = 1 + f[i + 1][j][0];
      ans = ans + f[i][j][0] + f[i][j][1] - 1;
      g[i][j] = 1;
    }
  }
  while (q--) {
    int l, r;
    cin >> l >> r;
    g[l][r] = !g[l][r];
    if (g[l][r]) {
      f[l][r][0] = 1 + f[l][r + 1][1];
      f[l][r][1] = 1 + f[l + 1][r][0];
      ans = ans + f[l][r][0] + f[l][r][1] - 1;
    } else {
      ans = ans - (f[l][r][0] + f[l][r][1] - 1);
      f[l][r][0] = f[l][r][1] = 0;
    }
    work(l, r - 1, 0);
    work(l - 1, r, 1);
    cout << ans << endl;
  }
  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;
const int inf = 0x3f3f3f3f;
const int dx[] = {0, 0, 1, -1}, dy[] = {1, -1, 0, 0};
void debug(int x = 0) { cout << "--ok(" << x << ")" << endl; }
const int N = 1e3 + 5;
long long f[N][N][2];
int g[N][N];
long long ans = 0, tot;
void work(int l, int r, int k) {
  if (l < 1 || r < 1) return;
  if (g[l][r] == 0) return;
  long long now = f[l][r][k];
  if (k == 0) {
    f[l][r][k] = f[l][r + 1][1] + 1;
    ans = ans + f[l][r][k] - now;
    work(l - 1, r, 1);
  } else {
    f[l][r][k] = f[l + 1][r][0] + 1;
    ans = ans + f[l][r][k] - now;
    work(l, r - 1, 0);
  }
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios::sync_with_stdio(false);
  int n, m, q;
  cin >> n >> m >> q;
  for (int i = n; i >= 1; i--) {
    for (int j = m; j >= 1; j--) {
      f[i][j][0] = 1 + f[i][j + 1][1];
      f[i][j][1] = 1 + f[i + 1][j][0];
      ans = ans + f[i][j][0] + f[i][j][1] - 1;
      g[i][j] = 1;
    }
  }
  while (q--) {
    int l, r;
    cin >> l >> r;
    g[l][r] = !g[l][r];
    if (g[l][r]) {
      f[l][r][0] = 1 + f[l][r + 1][1];
      f[l][r][1] = 1 + f[l + 1][r][0];
      ans = ans + f[l][r][0] + f[l][r][1] - 1;
    } else {
      ans = ans - (f[l][r][0] + f[l][r][1] - 1);
      f[l][r][0] = f[l][r][1] = 0;
    }
    work(l, r - 1, 0);
    work(l - 1, r, 1);
    cout << ans << endl;
  }
  return 0;
}

```