#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e3 + 5;
const long long mod = 998244353;
const double eps = 1e-10;
int n, m, q;
long long dp[maxn][maxn][2];
int vis[maxn][maxn];
long long ans;
long long dfs(int x, int y, int sgn) {
  if (dp[x][y][sgn]) return dp[x][y][sgn];
  dp[x][y][sgn] = 1;
  if (!sgn) {
    if (y < m) dp[x][y][sgn] += dfs(x, y + 1, sgn ^ 1);
  } else {
    if (x < n) dp[x][y][sgn] += dfs(x + 1, y, sgn ^ 1);
  }
  return dp[x][y][sgn];
}
void update(int x, int y, int sgn) {
  if (!vis[x][y]) {
    ans -= dp[x][y][0] + dp[x][y][1] - 1;
    dp[x][y][0] = dp[x][y][1] = 1;
    if (y < m) dp[x][y][0] += dp[x][y + 1][1];
    if (x < n) dp[x][y][1] += dp[x + 1][y][0];
    ans += dp[x][y][0] + dp[x][y][1] - 1;
  }
  if (!sgn) {
    if (x > 1) update(x - 1, y, 1);
  } else {
    if (y > 1) update(x, y - 1, 0);
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  dfs(1, 1, 0);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      dfs(i, j, 0);
      dfs(i, j, 1);
      ans += dp[i][j][0] + dp[i][j][1] - 1;
    }
  }
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (vis[x][y]) {
      dp[x][y][0] = dp[x][y][1] = 1;
      if (y < m) dp[x][y][0] += dp[x][y + 1][1];
      if (x < n) dp[x][y][1] += dp[x + 1][y][0];
      ans += dp[x][y][0] + dp[x][y][1] - 1;
      if (y > 1) update(x, y - 1, 0);
      if (x > 1) update(x - 1, y, 1);
      vis[x][y] = 0;
    } else {
      ans -= dp[x][y][0] + dp[x][y][1] - 1;
      dp[x][y][0] = dp[x][y][1] = 0;
      if (y > 1) update(x, y - 1, 0);
      if (x > 1) update(x - 1, y, 1);
      vis[x][y] = 1;
    }
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 = 1e3 + 5;
const long long mod = 998244353;
const double eps = 1e-10;
int n, m, q;
long long dp[maxn][maxn][2];
int vis[maxn][maxn];
long long ans;
long long dfs(int x, int y, int sgn) {
  if (dp[x][y][sgn]) return dp[x][y][sgn];
  dp[x][y][sgn] = 1;
  if (!sgn) {
    if (y < m) dp[x][y][sgn] += dfs(x, y + 1, sgn ^ 1);
  } else {
    if (x < n) dp[x][y][sgn] += dfs(x + 1, y, sgn ^ 1);
  }
  return dp[x][y][sgn];
}
void update(int x, int y, int sgn) {
  if (!vis[x][y]) {
    ans -= dp[x][y][0] + dp[x][y][1] - 1;
    dp[x][y][0] = dp[x][y][1] = 1;
    if (y < m) dp[x][y][0] += dp[x][y + 1][1];
    if (x < n) dp[x][y][1] += dp[x + 1][y][0];
    ans += dp[x][y][0] + dp[x][y][1] - 1;
  }
  if (!sgn) {
    if (x > 1) update(x - 1, y, 1);
  } else {
    if (y > 1) update(x, y - 1, 0);
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  dfs(1, 1, 0);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      dfs(i, j, 0);
      dfs(i, j, 1);
      ans += dp[i][j][0] + dp[i][j][1] - 1;
    }
  }
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (vis[x][y]) {
      dp[x][y][0] = dp[x][y][1] = 1;
      if (y < m) dp[x][y][0] += dp[x][y + 1][1];
      if (x < n) dp[x][y][1] += dp[x + 1][y][0];
      ans += dp[x][y][0] + dp[x][y][1] - 1;
      if (y > 1) update(x, y - 1, 0);
      if (x > 1) update(x - 1, y, 1);
      vis[x][y] = 0;
    } else {
      ans -= dp[x][y][0] + dp[x][y][1] - 1;
      dp[x][y][0] = dp[x][y][1] = 0;
      if (y > 1) update(x, y - 1, 0);
      if (x > 1) update(x - 1, y, 1);
      vis[x][y] = 1;
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```