#include <bits/stdc++.h>
using namespace std;
clock_t time_p = clock();
void time_taken() {
  time_p = clock() - time_p;
  cerr << "Time Taken : " << (float)(time_p) / CLOCKS_PER_SEC << "\n";
}
const long long mod = 1e9 + 7;
const long long INF = 1e18;
const int N = 1004;
int dp[N][N][2];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  int n, m, q;
  cin >> n >> m >> q;
  vector<vector<int> > a(n, vector<int>(m));
  memset(dp, 0, sizeof(dp));
  dp[n - 1][m - 1][1] = 1;
  dp[n - 1][m - 1][0] = 1;
  int ans = 0;
  for (int i = n - 1; i >= 0; i--) {
    for (int j = m - 1; j >= 0; j--) {
      dp[i][j][0] = 1;
      if (j + 1 < m) dp[i][j][0] += dp[i][j + 1][1];
      dp[i][j][1] = 1;
      if (i + 1 < n) dp[i][j][1] += dp[i + 1][j][0];
      ans += (dp[i][j][0] + dp[i][j][1] - 1);
    }
  };
  auto upd = [&](int r, int c, int delta) {
    if (delta == 0) return;
    ans += delta;
    dp[r][c][0] += delta;
    r--;
    while (r >= 0 and c >= 0) {
      if (a[r][c] == 1) return;
      dp[r][c][1] += delta;
      ans += delta;
      if (c) {
        if (a[r][c - 1] == 1) return;
        dp[r][c - 1][0] += delta;
        ans += delta;
      }
      c--;
      r--;
    }
  };
  while (q--) {
    int r, c;
    cin >> r >> c, --r, --c;
    a[r][c] ^= 1;
    int delta = 0;
    if (a[r][c] == 1) {
      delta -= 1;
      if (c + 1 < m) delta -= dp[r][c + 1][1];
    } else {
      delta += 1;
      if (c + 1 < m) delta += dp[r][c + 1][1];
    }
    upd(r, c, delta);
    delta = 0;
    if (a[r][c] == 1) {
      delta -= 1;
      if (r + 1 < n) delta -= dp[r + 1][c][0];
    } else {
      delta += 1;
      if (r + 1 < n) delta += dp[r + 1][c][0];
    }
    dp[r][c][1] += delta;
    ans += delta;
    if (c > 0 and a[r][c - 1] == 0) {
      upd(r, c - 1, delta);
    }
    if (a[r][c] == 1) {
      ans++;
    } else
      ans--;
    cout << ans << '\n';
  }
  time_taken();
  return 0;
}

### Prompt

Please create a solution in cpp 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;
clock_t time_p = clock();
void time_taken() {
  time_p = clock() - time_p;
  cerr << "Time Taken : " << (float)(time_p) / CLOCKS_PER_SEC << "\n";
}
const long long mod = 1e9 + 7;
const long long INF = 1e18;
const int N = 1004;
int dp[N][N][2];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  int n, m, q;
  cin >> n >> m >> q;
  vector<vector<int> > a(n, vector<int>(m));
  memset(dp, 0, sizeof(dp));
  dp[n - 1][m - 1][1] = 1;
  dp[n - 1][m - 1][0] = 1;
  int ans = 0;
  for (int i = n - 1; i >= 0; i--) {
    for (int j = m - 1; j >= 0; j--) {
      dp[i][j][0] = 1;
      if (j + 1 < m) dp[i][j][0] += dp[i][j + 1][1];
      dp[i][j][1] = 1;
      if (i + 1 < n) dp[i][j][1] += dp[i + 1][j][0];
      ans += (dp[i][j][0] + dp[i][j][1] - 1);
    }
  };
  auto upd = [&](int r, int c, int delta) {
    if (delta == 0) return;
    ans += delta;
    dp[r][c][0] += delta;
    r--;
    while (r >= 0 and c >= 0) {
      if (a[r][c] == 1) return;
      dp[r][c][1] += delta;
      ans += delta;
      if (c) {
        if (a[r][c - 1] == 1) return;
        dp[r][c - 1][0] += delta;
        ans += delta;
      }
      c--;
      r--;
    }
  };
  while (q--) {
    int r, c;
    cin >> r >> c, --r, --c;
    a[r][c] ^= 1;
    int delta = 0;
    if (a[r][c] == 1) {
      delta -= 1;
      if (c + 1 < m) delta -= dp[r][c + 1][1];
    } else {
      delta += 1;
      if (c + 1 < m) delta += dp[r][c + 1][1];
    }
    upd(r, c, delta);
    delta = 0;
    if (a[r][c] == 1) {
      delta -= 1;
      if (r + 1 < n) delta -= dp[r + 1][c][0];
    } else {
      delta += 1;
      if (r + 1 < n) delta += dp[r + 1][c][0];
    }
    dp[r][c][1] += delta;
    ans += delta;
    if (c > 0 and a[r][c - 1] == 0) {
      upd(r, c - 1, delta);
    }
    if (a[r][c] == 1) {
      ans++;
    } else
      ans--;
    cout << ans << '\n';
  }
  time_taken();
  return 0;
}

```