#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double PI = acos(-1.0);
int dp[1111][1111][2];
int mx[2] = {0, 1};
int my[2] = {1, 0};
struct node {
  int x, y, step, dir, type;
};
int n, m, q;
int mp[1111][1111];
int bfs(int x, int y, int type, int dir) {
  int step = 1;
  while (1) {
    if (type)
      y += dir;
    else
      x += dir;
    if (x >= 1 && x <= n && y >= 1 && y <= m && !mp[x][y])
      step++, type ^= 1;
    else
      return step;
  }
}
void solve() {
  cin >> n >> m >> q;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      dp[i][j][0] = dp[i][j - 1][1] + 1;
      dp[i][j][1] = dp[i - 1][j][0] + 1;
      ans += dp[i][j][0] + dp[i][j][1] - 1;
    }
  }
  while (q--) {
    int x, y;
    scanf("%d %d", &x, &y);
    int cnt1 = bfs(x, y, 1, 1), cnt2 = bfs(x, y, 0, -1);
    int cnt3 = bfs(x, y, 0, 1), cnt4 = bfs(x, y, 1, -1);
    mp[x][y] ^= 1;
    int now = cnt1 * cnt2 + cnt3 * cnt4 - 1;
    if (mp[x][y])
      ans -= now;
    else
      ans += now;
    printf("%lld\n", ans);
  }
}
int main(int argc, char const *argv[]) {
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### 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 int maxn = 1e5 + 10;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double PI = acos(-1.0);
int dp[1111][1111][2];
int mx[2] = {0, 1};
int my[2] = {1, 0};
struct node {
  int x, y, step, dir, type;
};
int n, m, q;
int mp[1111][1111];
int bfs(int x, int y, int type, int dir) {
  int step = 1;
  while (1) {
    if (type)
      y += dir;
    else
      x += dir;
    if (x >= 1 && x <= n && y >= 1 && y <= m && !mp[x][y])
      step++, type ^= 1;
    else
      return step;
  }
}
void solve() {
  cin >> n >> m >> q;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      dp[i][j][0] = dp[i][j - 1][1] + 1;
      dp[i][j][1] = dp[i - 1][j][0] + 1;
      ans += dp[i][j][0] + dp[i][j][1] - 1;
    }
  }
  while (q--) {
    int x, y;
    scanf("%d %d", &x, &y);
    int cnt1 = bfs(x, y, 1, 1), cnt2 = bfs(x, y, 0, -1);
    int cnt3 = bfs(x, y, 0, 1), cnt4 = bfs(x, y, 1, -1);
    mp[x][y] ^= 1;
    int now = cnt1 * cnt2 + cnt3 * cnt4 - 1;
    if (mp[x][y])
      ans -= now;
    else
      ans += now;
    printf("%lld\n", ans);
  }
}
int main(int argc, char const *argv[]) {
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```