#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
using PP = pair<ll, P>;
const ll n_ = 2222 + 100, inf = 1e18 + 1, mod = 1e9 + 7, sqrtN = 333;
ll dy[4] = {-1, 0, 1, 0}, dx[4] = {0, 1, 0, -1};
ll n, m, k, tc = 1, a, b, c, sum, x, y, z, w, base, ans;
ll gcd(ll x, ll y) {
  if (!y) return x;
  return gcd(y, x % y);
}
ll dp[n_][n_][2], A[n_][n_];
void solve() {
  ll q;
  cin >> n >> m >> q;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) A[i][j] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      dp[i][j][0] = A[i][j] * (dp[i][j - 1][1] + A[i][j - 1]);
      dp[i][j][1] = A[i][j] * (dp[i - 1][j][0] + A[i - 1][j]);
      ans += dp[i][j][0] + dp[i][j][1] + A[i][j];
    }
  ll i, j;
  while (q--) {
    cin >> i >> j;
    ans -= dp[i][j][0] + dp[i][j][1] + A[i][j];
    A[i][j] ^= 1;
    dp[i][j][0] = A[i][j] * (dp[i][j - 1][1] + A[i][j - 1]);
    dp[i][j][1] = A[i][j] * (dp[i - 1][j][0] + A[i - 1][j]);
    ans += dp[i][j][0] + dp[i][j][1] + A[i][j];
    b = i + 1, a = j, c = 0;
    while (1) {
      if (b > n || a > m) break;
      ans -= dp[b][a][0] + dp[b][a][1] + A[b][a];
      dp[b][a][0] = A[b][a] * (dp[b][a - 1][1] + A[b][a - 1]);
      dp[b][a][1] = A[b][a] * (dp[b - 1][a][0] + A[b - 1][a]);
      ans += dp[b][a][0] + dp[b][a][1] + A[b][a];
      if (c)
        b++;
      else
        a++;
      c = (c + 1) % 2;
    }
    b = i, a = j + 1, c = 1;
    while (1) {
      if (b > n || a > m) break;
      ans -= dp[b][a][0] + dp[b][a][1] + A[b][a];
      dp[b][a][0] = A[b][a] * (dp[b][a - 1][1] + A[b][a - 1]);
      dp[b][a][1] = A[b][a] * (dp[b - 1][a][0] + A[b - 1][a]);
      ans += dp[b][a][0] + dp[b][a][1] + A[b][a];
      if (c)
        b++;
      else
        a++;
      c = (c + 1) % 2;
    }
    cout << ans << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  while (tc--) solve();
}

### Prompt

Please formulate a cpp solution 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;
using ll = long long;
using P = pair<ll, ll>;
using PP = pair<ll, P>;
const ll n_ = 2222 + 100, inf = 1e18 + 1, mod = 1e9 + 7, sqrtN = 333;
ll dy[4] = {-1, 0, 1, 0}, dx[4] = {0, 1, 0, -1};
ll n, m, k, tc = 1, a, b, c, sum, x, y, z, w, base, ans;
ll gcd(ll x, ll y) {
  if (!y) return x;
  return gcd(y, x % y);
}
ll dp[n_][n_][2], A[n_][n_];
void solve() {
  ll q;
  cin >> n >> m >> q;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) A[i][j] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      dp[i][j][0] = A[i][j] * (dp[i][j - 1][1] + A[i][j - 1]);
      dp[i][j][1] = A[i][j] * (dp[i - 1][j][0] + A[i - 1][j]);
      ans += dp[i][j][0] + dp[i][j][1] + A[i][j];
    }
  ll i, j;
  while (q--) {
    cin >> i >> j;
    ans -= dp[i][j][0] + dp[i][j][1] + A[i][j];
    A[i][j] ^= 1;
    dp[i][j][0] = A[i][j] * (dp[i][j - 1][1] + A[i][j - 1]);
    dp[i][j][1] = A[i][j] * (dp[i - 1][j][0] + A[i - 1][j]);
    ans += dp[i][j][0] + dp[i][j][1] + A[i][j];
    b = i + 1, a = j, c = 0;
    while (1) {
      if (b > n || a > m) break;
      ans -= dp[b][a][0] + dp[b][a][1] + A[b][a];
      dp[b][a][0] = A[b][a] * (dp[b][a - 1][1] + A[b][a - 1]);
      dp[b][a][1] = A[b][a] * (dp[b - 1][a][0] + A[b - 1][a]);
      ans += dp[b][a][0] + dp[b][a][1] + A[b][a];
      if (c)
        b++;
      else
        a++;
      c = (c + 1) % 2;
    }
    b = i, a = j + 1, c = 1;
    while (1) {
      if (b > n || a > m) break;
      ans -= dp[b][a][0] + dp[b][a][1] + A[b][a];
      dp[b][a][0] = A[b][a] * (dp[b][a - 1][1] + A[b][a - 1]);
      dp[b][a][1] = A[b][a] * (dp[b - 1][a][0] + A[b - 1][a]);
      ans += dp[b][a][0] + dp[b][a][1] + A[b][a];
      if (c)
        b++;
      else
        a++;
      c = (c + 1) % 2;
    }
    cout << ans << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  while (tc--) solve();
}

```