#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long long n, m, q, dp[1003][1003][4], did[1003][1003], ans, num, xx, yy;
int go(int x, int y, int wt, int ox, int oy) {
  if (did[x][y] && (x != ox || y != oy)) return 0;
  if (x == n || y == m) return 0;
  if (wt == 0) return go(x + 1, y, 1, ox, oy) + 1;
  if (wt == 1) return go(x, y + 1, 0, ox, oy) + 1;
}
void rgo(int x, int y, int wt, int ox, int oy, int hm) {
  if (x == -1 || y == -1) return;
  if (did[x][y] && (x != ox || y != oy)) {
    if (wt == 1) {
      if (did[ox][oy])
        ans -= hm * dp[x][y][1];
      else
        ans += hm * dp[x][y][1];
    } else {
      if (did[ox][oy])
        ans -= hm * dp[x][y][0];
      else
        ans += hm * dp[x][y][0];
    }
    return;
  }
  if (wt == 0) rgo(x - 1, y, 1, ox, oy, hm);
  if (wt == 1) rgo(x, y - 1, 0, ox, oy, hm);
}
int main() {
  cin >> n >> m >> q;
  num = n * m;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      dp[i][j][0] = 1;
      dp[i][j][1] = 1;
      if (i - 1 >= 0) dp[i][j][0] += dp[i - 1][j][1];
      if (j - 1 >= 0) dp[i][j][1] += dp[i][j - 1][0];
      ans += dp[i][j][0];
      ans += dp[i][j][1];
    }
  }
  while (q--) {
    int x, y;
    cin >> x >> y;
    x--, y--;
    long long hm = go(x, y, 0, x, y);
    if (did[x][y])
      ans += hm * dp[x][y][1];
    else
      ans -= hm * dp[x][y][1];
    rgo(x, y, 1, x, y, hm);
    hm = go(x, y, 1, x, y);
    if (did[x][y])
      ans += hm * dp[x][y][0];
    else
      ans -= hm * dp[x][y][0];
    rgo(x, y, 0, x, y, hm);
    if (did[x][y])
      num++;
    else
      num--;
    did[x][y] ^= 1;
    cout << ans - num << endl;
  }
}

### Prompt

In cpp, your task is to solve 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 mod = 1e9 + 7;
long long n, m, q, dp[1003][1003][4], did[1003][1003], ans, num, xx, yy;
int go(int x, int y, int wt, int ox, int oy) {
  if (did[x][y] && (x != ox || y != oy)) return 0;
  if (x == n || y == m) return 0;
  if (wt == 0) return go(x + 1, y, 1, ox, oy) + 1;
  if (wt == 1) return go(x, y + 1, 0, ox, oy) + 1;
}
void rgo(int x, int y, int wt, int ox, int oy, int hm) {
  if (x == -1 || y == -1) return;
  if (did[x][y] && (x != ox || y != oy)) {
    if (wt == 1) {
      if (did[ox][oy])
        ans -= hm * dp[x][y][1];
      else
        ans += hm * dp[x][y][1];
    } else {
      if (did[ox][oy])
        ans -= hm * dp[x][y][0];
      else
        ans += hm * dp[x][y][0];
    }
    return;
  }
  if (wt == 0) rgo(x - 1, y, 1, ox, oy, hm);
  if (wt == 1) rgo(x, y - 1, 0, ox, oy, hm);
}
int main() {
  cin >> n >> m >> q;
  num = n * m;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      dp[i][j][0] = 1;
      dp[i][j][1] = 1;
      if (i - 1 >= 0) dp[i][j][0] += dp[i - 1][j][1];
      if (j - 1 >= 0) dp[i][j][1] += dp[i][j - 1][0];
      ans += dp[i][j][0];
      ans += dp[i][j][1];
    }
  }
  while (q--) {
    int x, y;
    cin >> x >> y;
    x--, y--;
    long long hm = go(x, y, 0, x, y);
    if (did[x][y])
      ans += hm * dp[x][y][1];
    else
      ans -= hm * dp[x][y][1];
    rgo(x, y, 1, x, y, hm);
    hm = go(x, y, 1, x, y);
    if (did[x][y])
      ans += hm * dp[x][y][0];
    else
      ans -= hm * dp[x][y][0];
    rgo(x, y, 0, x, y, hm);
    if (did[x][y])
      num++;
    else
      num--;
    did[x][y] ^= 1;
    cout << ans - num << endl;
  }
}

```