#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 10;
long long dp[N][N][2];
int vis[N][N];
pair<int, int> d[] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
pair<int, int> add(pair<int, int> a, pair<int, int> b) {
  return make_pair(a.first + b.first, a.second + b.second);
}
int main() {
  long long n, m, q;
  scanf("%lld%lld%lld", &n, &m, &q);
  long long ans = 0;
  memset(vis, -1, sizeof(vis));
  for (int i = n; i >= 1; i--)
    for (int j = m; j >= 1; j--) {
      vis[i][j] = 0;
      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--) {
    long long x, y;
    scanf("%lld%lld", &x, &y);
    vector<long long> v;
    for (int i = 0; i < 4; i++) {
      pair<int, int> now = make_pair(x, y);
      long long len = 1, id = i;
      while (1) {
        now = add(now, d[id]);
        id ^= 1;
        if (vis[now.first][now.second]) break;
        len++;
      }
      v.push_back(len);
    }
    long long tmp = v[0] * v[3] + v[1] * v[2] - 1;
    ans += (vis[x][y] ? tmp : -tmp);
    vis[x][y] ^= 1;
    printf("%lld\n", ans);
  }
}

### Prompt

Develop a solution in CPP to the problem described below:

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 N = 1e3 + 10;
long long dp[N][N][2];
int vis[N][N];
pair<int, int> d[] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
pair<int, int> add(pair<int, int> a, pair<int, int> b) {
  return make_pair(a.first + b.first, a.second + b.second);
}
int main() {
  long long n, m, q;
  scanf("%lld%lld%lld", &n, &m, &q);
  long long ans = 0;
  memset(vis, -1, sizeof(vis));
  for (int i = n; i >= 1; i--)
    for (int j = m; j >= 1; j--) {
      vis[i][j] = 0;
      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--) {
    long long x, y;
    scanf("%lld%lld", &x, &y);
    vector<long long> v;
    for (int i = 0; i < 4; i++) {
      pair<int, int> now = make_pair(x, y);
      long long len = 1, id = i;
      while (1) {
        now = add(now, d[id]);
        id ^= 1;
        if (vis[now.first][now.second]) break;
        len++;
      }
      v.push_back(len);
    }
    long long tmp = v[0] * v[3] + v[1] * v[2] - 1;
    ans += (vis[x][y] ? tmp : -tmp);
    vis[x][y] ^= 1;
    printf("%lld\n", ans);
  }
}

```