Pawns
Contest: Educational Codeforces Round 90 (Rated for Div. 2)
Difficulty rating: 2600
Tags: data structures, divide and conquer, greedy

You are given a chessboard consisting of $$$n$$$ rows and $$$n$$$ columns. Rows are numbered from bottom to top from $$$1$$$ to $$$n$$$. Columns are numbered from left to right from $$$1$$$ to $$$n$$$. The cell at the intersection of the $$$x$$$-th column and the $$$y$$$-th row is denoted as $$$(x, y)$$$. Furthermore, the $$$k$$$-th column is a special column.

Initially, the board is empty. There are $$$m$$$ changes to the board. During the $$$i$$$-th change one pawn is added or removed from the board. The current board is good if we can move all pawns to the special column by the followings rules:

- Pawn in the cell $$$(x, y)$$$ can be moved to the cell $$$(x, y + 1)$$$, $$$(x - 1, y + 1)$$$ or $$$(x + 1, y + 1)$$$;
- You can make as many such moves as you like;
- Pawns can not be moved outside the chessboard;
- Each cell can not contain more than one pawn.

The current board may not always be good. To fix it, you can add new rows to the board. New rows are added at the top, i. e. they will have numbers $$$n+1, n+2, n+3, \dots$$$.

After each of $$$m$$$ changes, print one integer — the minimum number of rows which you have to add to make the board good.

Input format:
The first line contains three integers $$$n$$$, $$$k$$$ and $$$m$$$ ($$$1 \le n, m \le 2 \cdot 10^5; 1 \le k \le n$$$) — the size of the board, the index of the special column and the number of changes respectively.

Then $$$m$$$ lines follow. The $$$i$$$-th line contains two integers $$$x$$$ and $$$y$$$ ($$$1 \le x, y \le n$$$) — the index of the column and the index of the row respectively. If there is no pawn in the cell $$$(x, y)$$$, then you add a pawn to this cell, otherwise — you remove the pawn from this cell.

Output format:
After each change print one integer — the minimum number of rows which you have to add to make the board good.

Constraints:
Time limit: 3.0 s
Memory limit: 256.0 MB

Example input:
5 3 5
4 4
3 5
2 4
3 4
3 5

Example output:
0
1
2
2
1

Editorial:
For each pawn with initial position $$$(x, y)$$$ there exists a minimum index of row $$$i$$$ such that the pawn can reach the cell $$$(k, i)$$$, but cannot reach the cell $$$(k, i - 1)$$$. It's easy to see that $$$i = |k - x| + y$$$. In the resulting configuration, this pawn can occupy the cell $$$(k, i)$$$, $$$(k, i + 1)$$$, $$$(k, i + 2)$$$ or any other cell $$$(k, j)$$$ having $$$j \ge i$$$.
Suppose the board consists of $$$r$$$ rows. For each row, the number of rows above it should be not less than the number of pawns that occupy the cells above it (that is, having $$$i$$$ greater than the index of that row) — because, if this condition is not fulfilled, we can't assign each pawn a unique cell. If we denote the number of pawns that should go strictly above the $$$j$$$-th row as $$$f(j)$$$, then for every row, the condition $$$f(j) \le r - j$$$ must be met.
To prove that this condition is sufficient, we may, for example, use Hall's theorem. Okay, now what about finding the minimum $$$r$$$ satisfying it? Let's initially set $$$r$$$ to $$$n$$$, and for each row maintain the value of $$$j + f(j) - n$$$ — the minimum number of rows we have to add to our board so that the condition for the row $$$j$$$ is met (we also have to maintain this value for $$$n - 1$$$ auxiliary rows from $$$n + 1$$$ to $$$2n - 1$$$, since some pawns cannot fit in the initial board at all). Finding the minimum value we have to add to $$$r$$$ equals finding the maximum of all these values on some prefix (we don't need to look at the values on some rows with large indices, if there are no pawns after them, so we need a maximum query on the segment $$$[1, \max_i]$$$, where $$$\max_i$$$ is the maximum index $$$i$$$ among all pawns); and when a pawn is added or removed, we should add $$$+1$$$ or $$$-1$$$ to all values on some suffix. A segment tree with lazy propagation will do the trick, solving the problem for us in $$$O(m \log n)$$$.