#include <bits/stdc++.h>
using namespace std;
long long a[1010][1010][2];
long long n, m, q;
long long col[1010][1010];
long long sum = 0;
long long cntCl = 0;
void build() {
  for (long long i = 1; i <= n; i++) {
    for (long long i1 = 1; i1 <= m; i1++) {
      long long right1 = m - i1, right2 = n - i;
      if (right1 <= right2) {
        a[i][i1][0] = right1 * 2;
      } else {
        a[i][i1][0] = 1 + right2 * 2;
      }
      long long bottom1 = m - i1, bottom2 = n - i;
      if (bottom1 < bottom2) {
        a[i][i1][1] = 1 + bottom1 * 2;
      } else {
        a[i][i1][1] = bottom2 * 2;
      }
      sum += (a[i][i1][0] + a[i][i1][1]);
    }
  }
}
long long calc_new(long long x, long long y, long long nxt) {
  if (x > n || y > m || col[x][y] == 1) {
    return 0;
  }
  return 1 + calc_new(x + (nxt == 1), y + (nxt == 0), 1 - nxt);
}
void rise_up(long long x, long long y, long long val, long long nxt) {
  if (x < 1 || y < 1 || col[x][y] == 1) {
    return;
  }
  sum += val;
  a[x][y][1 - nxt] += val;
  rise_up(x - (nxt == 1), y - (nxt == 0), val, 1 - nxt);
}
void printA() {
  cout << "------------------->\n";
  for (long long i = 1; i <= n; i++) {
    for (long long i1 = 1; i1 <= m; i1++) {
      cout << a[i][i1][1] << " ";
    }
    cout << endl;
  }
  cout << "<-------------------\n";
}
void solve() {
  cin >> n >> m >> q;
  build();
  cntCl = n * m;
  for (long long i = 1; i <= q; i++) {
    long long x, y;
    cin >> x >> y;
    {
      long long cnt_add = 0;
      if (col[x][y] == 0) {
        cnt_add = calc_new(x, y, 0);
        rise_up(x, y, -(cnt_add), 1);
        sum++;
        a[x][y][1]++;
      } else {
        col[x][y] = 0;
        cnt_add = calc_new(x, y, 0);
        rise_up(x, y, (cnt_add), 1);
        col[x][y] = 1;
        sum--;
        a[x][y][1]--;
      }
    }
    {
      long long cnt_add = 0;
      if (col[x][y] == 0) {
        cnt_add = calc_new(x, y, 1);
        rise_up(x, y, -(cnt_add), 0);
        sum++;
        a[x][y][0]++;
      } else {
        col[x][y] = 0;
        cnt_add = calc_new(x, y, 1);
        rise_up(x, y, (cnt_add), 0);
        col[x][y] = 1;
        sum--;
        a[x][y][0]--;
      }
    }
    cntCl += (col[x][y] == 1 ? 1 : -1);
    col[x][y] = 1 - col[x][y];
    cout << sum + cntCl << "\n";
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long q = 1;
  for (long long i = 1; i <= q; i++) {
    solve();
  }
}

### Prompt

In cpp, your task is to solve 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;
long long a[1010][1010][2];
long long n, m, q;
long long col[1010][1010];
long long sum = 0;
long long cntCl = 0;
void build() {
  for (long long i = 1; i <= n; i++) {
    for (long long i1 = 1; i1 <= m; i1++) {
      long long right1 = m - i1, right2 = n - i;
      if (right1 <= right2) {
        a[i][i1][0] = right1 * 2;
      } else {
        a[i][i1][0] = 1 + right2 * 2;
      }
      long long bottom1 = m - i1, bottom2 = n - i;
      if (bottom1 < bottom2) {
        a[i][i1][1] = 1 + bottom1 * 2;
      } else {
        a[i][i1][1] = bottom2 * 2;
      }
      sum += (a[i][i1][0] + a[i][i1][1]);
    }
  }
}
long long calc_new(long long x, long long y, long long nxt) {
  if (x > n || y > m || col[x][y] == 1) {
    return 0;
  }
  return 1 + calc_new(x + (nxt == 1), y + (nxt == 0), 1 - nxt);
}
void rise_up(long long x, long long y, long long val, long long nxt) {
  if (x < 1 || y < 1 || col[x][y] == 1) {
    return;
  }
  sum += val;
  a[x][y][1 - nxt] += val;
  rise_up(x - (nxt == 1), y - (nxt == 0), val, 1 - nxt);
}
void printA() {
  cout << "------------------->\n";
  for (long long i = 1; i <= n; i++) {
    for (long long i1 = 1; i1 <= m; i1++) {
      cout << a[i][i1][1] << " ";
    }
    cout << endl;
  }
  cout << "<-------------------\n";
}
void solve() {
  cin >> n >> m >> q;
  build();
  cntCl = n * m;
  for (long long i = 1; i <= q; i++) {
    long long x, y;
    cin >> x >> y;
    {
      long long cnt_add = 0;
      if (col[x][y] == 0) {
        cnt_add = calc_new(x, y, 0);
        rise_up(x, y, -(cnt_add), 1);
        sum++;
        a[x][y][1]++;
      } else {
        col[x][y] = 0;
        cnt_add = calc_new(x, y, 0);
        rise_up(x, y, (cnt_add), 1);
        col[x][y] = 1;
        sum--;
        a[x][y][1]--;
      }
    }
    {
      long long cnt_add = 0;
      if (col[x][y] == 0) {
        cnt_add = calc_new(x, y, 1);
        rise_up(x, y, -(cnt_add), 0);
        sum++;
        a[x][y][0]++;
      } else {
        col[x][y] = 0;
        cnt_add = calc_new(x, y, 1);
        rise_up(x, y, (cnt_add), 0);
        col[x][y] = 1;
        sum--;
        a[x][y][0]--;
      }
    }
    cntCl += (col[x][y] == 1 ? 1 : -1);
    col[x][y] = 1 - col[x][y];
    cout << sum + cntCl << "\n";
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long q = 1;
  for (long long i = 1; i <= q; i++) {
    solve();
  }
}

```