#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long n, m, q;
  cin >> n >> m >> q;
  vector<vector<bool>> a(n, vector<bool>(m, 1));
  long long total = 0;
  vector<vector<pair<long long, long long>>> dp(
      n, vector<pair<long long, long long>>(m));
  for (long long i = 0; i < n; ++i) {
    for (long long j = 0; j < m; ++j) {
      if (i == 0) {
        if (j == 0)
          ;
        else
          dp[i][j].second = 1;
      } else if (j == 0)
        dp[i][j].first = 1;
      else
        dp[i][j].first = (dp[i - 1][j].second + 1),
        dp[i][j].second = (dp[i][j - 1].first + 1);
      total += (1 + dp[i][j].first + dp[i][j].second);
    }
  }
  while (q--) {
    long long r, c;
    cin >> r >> c;
    --r, --c;
    long long cur_c = c, cur_r = r;
    bool add = 0;
    if (!a[r][c]) add = 1;
    long long ts = 0, trn = 0, t1u = 0, t2u = 0, t1d = 0, t2d = 0;
    while (cur_c >= 0 && cur_r >= 0) {
      if (cur_r == r && cur_c == c) {
        t1u += 1;
        cur_r -= 1;
        trn++;
        continue;
      }
      if (a[cur_r][cur_c] == 0) break;
      if (trn == 0)
        cur_r -= 1;
      else
        cur_c -= 1;
      trn++;
      trn %= 2;
      t1u++;
    }
    cur_r = r, cur_c = c;
    trn = 0;
    while (cur_r >= 0 && cur_c >= 0) {
      if (cur_r == r && cur_c == c) {
        cur_c -= 1;
        trn++;
        continue;
      }
      if (a[cur_r][cur_c] == 0) break;
      if (trn == 0)
        cur_c -= 1;
      else
        cur_r -= 1;
      trn++;
      trn %= 2;
      t2u++;
    }
    cur_r = r, cur_c = c;
    trn = 0;
    while (cur_r < n && cur_c < m) {
      if (cur_r == r && cur_c == c) {
        cur_r += 1;
        trn++;
        continue;
      }
      if (a[cur_r][cur_c] == 0) break;
      if (trn == 0)
        cur_r += 1;
      else
        cur_c += 1;
      trn++;
      trn %= 2;
      t1d++;
    }
    cur_r = r, cur_c = c;
    trn = 0;
    while (cur_r < n && cur_c < m) {
      if (cur_r == r && cur_c == c) {
        cur_c += 1;
        trn++;
        continue;
      }
      if (a[cur_r][cur_c] == 0) break;
      ;
      if (trn == 0)
        cur_c += 1;
      else
        cur_r += 1;
      trn++;
      trn %= 2;
      t2d++;
    }
    ts = (t1u + t2u + t1d + t2d);
    ts += ((t1u - 1) * t2d + t2u * t1d);
    a[r][c] = (!a[r][c]);
    if (add)
      total += ts;
    else
      total -= ts;
    cout << total << '\n';
  }
}
int32_t main() {
  long long nt;
  nt = 1;
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  while (nt--) {
    solve();
  }
}

### Prompt

Generate 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;
void solve() {
  long long n, m, q;
  cin >> n >> m >> q;
  vector<vector<bool>> a(n, vector<bool>(m, 1));
  long long total = 0;
  vector<vector<pair<long long, long long>>> dp(
      n, vector<pair<long long, long long>>(m));
  for (long long i = 0; i < n; ++i) {
    for (long long j = 0; j < m; ++j) {
      if (i == 0) {
        if (j == 0)
          ;
        else
          dp[i][j].second = 1;
      } else if (j == 0)
        dp[i][j].first = 1;
      else
        dp[i][j].first = (dp[i - 1][j].second + 1),
        dp[i][j].second = (dp[i][j - 1].first + 1);
      total += (1 + dp[i][j].first + dp[i][j].second);
    }
  }
  while (q--) {
    long long r, c;
    cin >> r >> c;
    --r, --c;
    long long cur_c = c, cur_r = r;
    bool add = 0;
    if (!a[r][c]) add = 1;
    long long ts = 0, trn = 0, t1u = 0, t2u = 0, t1d = 0, t2d = 0;
    while (cur_c >= 0 && cur_r >= 0) {
      if (cur_r == r && cur_c == c) {
        t1u += 1;
        cur_r -= 1;
        trn++;
        continue;
      }
      if (a[cur_r][cur_c] == 0) break;
      if (trn == 0)
        cur_r -= 1;
      else
        cur_c -= 1;
      trn++;
      trn %= 2;
      t1u++;
    }
    cur_r = r, cur_c = c;
    trn = 0;
    while (cur_r >= 0 && cur_c >= 0) {
      if (cur_r == r && cur_c == c) {
        cur_c -= 1;
        trn++;
        continue;
      }
      if (a[cur_r][cur_c] == 0) break;
      if (trn == 0)
        cur_c -= 1;
      else
        cur_r -= 1;
      trn++;
      trn %= 2;
      t2u++;
    }
    cur_r = r, cur_c = c;
    trn = 0;
    while (cur_r < n && cur_c < m) {
      if (cur_r == r && cur_c == c) {
        cur_r += 1;
        trn++;
        continue;
      }
      if (a[cur_r][cur_c] == 0) break;
      if (trn == 0)
        cur_r += 1;
      else
        cur_c += 1;
      trn++;
      trn %= 2;
      t1d++;
    }
    cur_r = r, cur_c = c;
    trn = 0;
    while (cur_r < n && cur_c < m) {
      if (cur_r == r && cur_c == c) {
        cur_c += 1;
        trn++;
        continue;
      }
      if (a[cur_r][cur_c] == 0) break;
      ;
      if (trn == 0)
        cur_c += 1;
      else
        cur_r += 1;
      trn++;
      trn %= 2;
      t2d++;
    }
    ts = (t1u + t2u + t1d + t2d);
    ts += ((t1u - 1) * t2d + t2u * t1d);
    a[r][c] = (!a[r][c]);
    if (add)
      total += ts;
    else
      total -= ts;
    cout << total << '\n';
  }
}
int32_t main() {
  long long nt;
  nt = 1;
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  while (nt--) {
    solve();
  }
}

```