#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pi = pair<ll, ll>;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const ll minf = 0xc0c0c0c0c0c0c0c0;
ll A[1010][1010];
ll dp[1010][1010], dp1[1010][1010], dp2[1010][1010];
ll res = 0;
void f(ll i, ll j) {
  if (A[i][j] == 1) {
    return;
  }
  dp[i][j] = dp1[i][j - 1] + dp2[i - 1][j] + 1;
  dp1[i][j] = dp2[i - 1][j] + 1;
  dp2[i][j] = dp1[i][j - 1] + 1;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ll n, m, Q;
  cin >> n >> m >> Q;
  for (ll i = 1; i <= n; i++) {
    for (ll j = 1; j <= m; j++) {
      f(i, j);
      res += dp[i][j];
    }
  }
  while (Q--) {
    ll x, y;
    cin >> x >> y;
    A[x][y] ^= 1;
    if (A[x][y] == 1) {
      res -= dp[x][y];
      dp[x][y] = dp1[x][y] = dp2[x][y] = 0;
      if (x < n) {
        res -= dp[x + 1][y];
        f(x + 1, y);
        res += dp[x + 1][y];
      }
      if (y < m) {
        res -= dp[x][y + 1];
        f(x, y + 1);
        res += dp[x][y + 1];
      }
      ++x, ++y;
      while (x <= n && y <= m) {
        res -= dp[x][y];
        f(x, y);
        res += dp[x][y];
        if (x < n) {
          res -= dp[x + 1][y];
          f(x + 1, y);
          res += dp[x + 1][y];
        }
        if (y < m) {
          res -= dp[x][y + 1];
          f(x, y + 1);
          res += dp[x][y + 1];
        }
        ++x, ++y;
      }
    } else {
      while (x <= n && y <= m) {
        res -= dp[x][y];
        f(x, y);
        res += dp[x][y];
        if (x < n) {
          res -= dp[x + 1][y];
          f(x + 1, y);
          res += dp[x + 1][y];
        }
        if (y < m) {
          res -= dp[x][y + 1];
          f(x, y + 1);
          res += dp[x][y + 1];
        }
        ++x, ++y;
      }
    }
    cout << res << "\n";
  }
  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;
using ll = long long;
using pi = pair<ll, ll>;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const ll minf = 0xc0c0c0c0c0c0c0c0;
ll A[1010][1010];
ll dp[1010][1010], dp1[1010][1010], dp2[1010][1010];
ll res = 0;
void f(ll i, ll j) {
  if (A[i][j] == 1) {
    return;
  }
  dp[i][j] = dp1[i][j - 1] + dp2[i - 1][j] + 1;
  dp1[i][j] = dp2[i - 1][j] + 1;
  dp2[i][j] = dp1[i][j - 1] + 1;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ll n, m, Q;
  cin >> n >> m >> Q;
  for (ll i = 1; i <= n; i++) {
    for (ll j = 1; j <= m; j++) {
      f(i, j);
      res += dp[i][j];
    }
  }
  while (Q--) {
    ll x, y;
    cin >> x >> y;
    A[x][y] ^= 1;
    if (A[x][y] == 1) {
      res -= dp[x][y];
      dp[x][y] = dp1[x][y] = dp2[x][y] = 0;
      if (x < n) {
        res -= dp[x + 1][y];
        f(x + 1, y);
        res += dp[x + 1][y];
      }
      if (y < m) {
        res -= dp[x][y + 1];
        f(x, y + 1);
        res += dp[x][y + 1];
      }
      ++x, ++y;
      while (x <= n && y <= m) {
        res -= dp[x][y];
        f(x, y);
        res += dp[x][y];
        if (x < n) {
          res -= dp[x + 1][y];
          f(x + 1, y);
          res += dp[x + 1][y];
        }
        if (y < m) {
          res -= dp[x][y + 1];
          f(x, y + 1);
          res += dp[x][y + 1];
        }
        ++x, ++y;
      }
    } else {
      while (x <= n && y <= m) {
        res -= dp[x][y];
        f(x, y);
        res += dp[x][y];
        if (x < n) {
          res -= dp[x + 1][y];
          f(x + 1, y);
          res += dp[x + 1][y];
        }
        if (y < m) {
          res -= dp[x][y + 1];
          f(x, y + 1);
          res += dp[x][y + 1];
        }
        ++x, ++y;
      }
    }
    cout << res << "\n";
  }
  return 0;
}

```