#include <bits/stdc++.h>
using namespace std;
const int N = 1009;
int vis[2][N][N], n, m, q, bad[N][N], vis2[N][N];
vector<int> mp[N][N];
void walk(vector<pair<int, int>> &staircase, int i, int j, int prev) {
  vis[prev][i][j] = 1;
  staircase.push_back({i, j});
  int nx = i + prev, ny = j + !prev;
  if (nx < n && ny < m) {
    assert(vis[!prev][nx][ny] == 0);
    walk(staircase, nx, ny, !prev);
  }
}
void solve() {
  cin >> n >> m >> q;
  vector<vector<pair<int, int>>> paths;
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      for (int k = 0; k < 2; k++) {
        if (!vis[k][i][j]) {
          if (!k && j == m - 1 && !(n == 1 && m == 1)) continue;
          if (k && i == n - 1 && !(n == 1 && m == 1)) continue;
          vector<pair<int, int>> v;
          walk(v, i, j, k);
          for (auto x : v) {
            mp[x.first][x.second].push_back(paths.size());
          }
          ans += (1ll * v.size() * (v.size() - 1)) / 2;
          paths.push_back(v);
        }
      }
    }
  }
  ans += n * m;
  while (q--) {
    int x, y;
    cin >> x >> y;
    for (int idx : mp[x - 1][y - 1]) {
      for (auto p : paths[idx]) {
        if (bad[p.first][p.second]) continue;
        if (!vis2[p.first][p.second]) {
          vis2[p.first][p.second] = 1;
          ans--;
        }
      }
      for (int i = 0; i < paths[idx].size();) {
        int ii = paths[idx][i].first;
        int jj = paths[idx][i].second;
        if (bad[ii][jj]) {
          i++;
          continue;
        }
        int j = i;
        while (j < paths[idx].size()) {
          ii = paths[idx][j].first;
          jj = paths[idx][j].second;
          if (bad[ii][jj]) break;
          j++;
        }
        int len = j - i;
        ans -= (1ll * len * (len - 1)) / 2;
        i = j;
      }
    }
    for (int idx : mp[x - 1][y - 1]) {
      for (auto p : paths[idx]) {
        if (bad[p.first][p.second]) continue;
        if (vis2[p.first][p.second]) {
          vis2[p.first][p.second] = 0;
        }
      }
    }
    bad[x - 1][y - 1] ^= 1;
    for (int idx : mp[x - 1][y - 1]) {
      for (auto p : paths[idx]) {
        if (bad[p.first][p.second]) continue;
        if (!vis2[p.first][p.second]) {
          vis2[p.first][p.second] = 1;
          ans++;
        }
      }
      for (int i = 0; i < paths[idx].size();) {
        int ii = paths[idx][i].first;
        int jj = paths[idx][i].second;
        if (bad[ii][jj]) {
          i++;
          continue;
        }
        int j = i;
        while (j < paths[idx].size()) {
          ii = paths[idx][j].first;
          jj = paths[idx][j].second;
          if (bad[ii][jj]) break;
          j++;
        }
        int len = j - i;
        ans += (1ll * len * (len - 1)) / 2;
        i = j;
      }
    }
    for (int idx : mp[x - 1][y - 1]) {
      for (auto p : paths[idx]) {
        if (bad[p.first][p.second]) continue;
        if (vis2[p.first][p.second]) {
          vis2[p.first][p.second] = 0;
        }
      }
    }
    cout << ans << "\n";
  }
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
}

### Prompt

Your challenge is to write 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;
const int N = 1009;
int vis[2][N][N], n, m, q, bad[N][N], vis2[N][N];
vector<int> mp[N][N];
void walk(vector<pair<int, int>> &staircase, int i, int j, int prev) {
  vis[prev][i][j] = 1;
  staircase.push_back({i, j});
  int nx = i + prev, ny = j + !prev;
  if (nx < n && ny < m) {
    assert(vis[!prev][nx][ny] == 0);
    walk(staircase, nx, ny, !prev);
  }
}
void solve() {
  cin >> n >> m >> q;
  vector<vector<pair<int, int>>> paths;
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      for (int k = 0; k < 2; k++) {
        if (!vis[k][i][j]) {
          if (!k && j == m - 1 && !(n == 1 && m == 1)) continue;
          if (k && i == n - 1 && !(n == 1 && m == 1)) continue;
          vector<pair<int, int>> v;
          walk(v, i, j, k);
          for (auto x : v) {
            mp[x.first][x.second].push_back(paths.size());
          }
          ans += (1ll * v.size() * (v.size() - 1)) / 2;
          paths.push_back(v);
        }
      }
    }
  }
  ans += n * m;
  while (q--) {
    int x, y;
    cin >> x >> y;
    for (int idx : mp[x - 1][y - 1]) {
      for (auto p : paths[idx]) {
        if (bad[p.first][p.second]) continue;
        if (!vis2[p.first][p.second]) {
          vis2[p.first][p.second] = 1;
          ans--;
        }
      }
      for (int i = 0; i < paths[idx].size();) {
        int ii = paths[idx][i].first;
        int jj = paths[idx][i].second;
        if (bad[ii][jj]) {
          i++;
          continue;
        }
        int j = i;
        while (j < paths[idx].size()) {
          ii = paths[idx][j].first;
          jj = paths[idx][j].second;
          if (bad[ii][jj]) break;
          j++;
        }
        int len = j - i;
        ans -= (1ll * len * (len - 1)) / 2;
        i = j;
      }
    }
    for (int idx : mp[x - 1][y - 1]) {
      for (auto p : paths[idx]) {
        if (bad[p.first][p.second]) continue;
        if (vis2[p.first][p.second]) {
          vis2[p.first][p.second] = 0;
        }
      }
    }
    bad[x - 1][y - 1] ^= 1;
    for (int idx : mp[x - 1][y - 1]) {
      for (auto p : paths[idx]) {
        if (bad[p.first][p.second]) continue;
        if (!vis2[p.first][p.second]) {
          vis2[p.first][p.second] = 1;
          ans++;
        }
      }
      for (int i = 0; i < paths[idx].size();) {
        int ii = paths[idx][i].first;
        int jj = paths[idx][i].second;
        if (bad[ii][jj]) {
          i++;
          continue;
        }
        int j = i;
        while (j < paths[idx].size()) {
          ii = paths[idx][j].first;
          jj = paths[idx][j].second;
          if (bad[ii][jj]) break;
          j++;
        }
        int len = j - i;
        ans += (1ll * len * (len - 1)) / 2;
        i = j;
      }
    }
    for (int idx : mp[x - 1][y - 1]) {
      for (auto p : paths[idx]) {
        if (bad[p.first][p.second]) continue;
        if (vis2[p.first][p.second]) {
          vis2[p.first][p.second] = 0;
        }
      }
    }
    cout << ans << "\n";
  }
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
}

```