#include <bits/stdc++.h>
using namespace std;
const int mxn = 1001;
int n, m, k, q;
int a[mxn][mxn], dp[mxn][mxn][2];
void first(int x, int y, int z) {
  k += z * (dp[x][y][0] + dp[x][y][1] - a[x][y]);
}
void ff(int x, int y, int z) {
  if (x > n || y > m) return;
  first(x, y, -1), a[x][y] ^= z;
  for (int i = 0; i < 2; i++)
    dp[x][y][i] = a[x][y] * (dp[x - !i][y - i][!i] + 1);
  first(x, y, 1);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> q;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      ff(i, j, 1);
    }
  while (q--) {
    int x, y;
    cin >> x >> y;
    ff(x, y, 1);
    for (x++, y++; x <= n + 1 && y <= m + 1; x++, y++) {
      ff(x, y - 1, 0), ff(x - 1, y, 0), ff(x, y, 0);
    }
    cout << k << '\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;
const int mxn = 1001;
int n, m, k, q;
int a[mxn][mxn], dp[mxn][mxn][2];
void first(int x, int y, int z) {
  k += z * (dp[x][y][0] + dp[x][y][1] - a[x][y]);
}
void ff(int x, int y, int z) {
  if (x > n || y > m) return;
  first(x, y, -1), a[x][y] ^= z;
  for (int i = 0; i < 2; i++)
    dp[x][y][i] = a[x][y] * (dp[x - !i][y - i][!i] + 1);
  first(x, y, 1);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> q;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      ff(i, j, 1);
    }
  while (q--) {
    int x, y;
    cin >> x >> y;
    ff(x, y, 1);
    for (x++, y++; x <= n + 1 && y <= m + 1; x++, y++) {
      ff(x, y - 1, 0), ff(x - 1, y, 0), ff(x, y, 0);
    }
    cout << k << '\n';
  }
  return 0;
}

```