Binary Table (Hard Version)
Contest: Codeforces Round 684 (Div. 1)
Difficulty rating: 1900
Tags: constructive algorithms, graphs, greedy, implementation

This is the hard version of the problem. The difference between the versions is in the number of possible operations that can be made. You can make hacks if and only if you solved both versions of the problem.

You are given a binary table of size $$$n \times m$$$. This table consists of symbols $$$0$$$ and $$$1$$$.

You can make such operation: select $$$3$$$ different cells that belong to one $$$2 \times 2$$$ square and change the symbols in these cells (change $$$0$$$ to $$$1$$$ and $$$1$$$ to $$$0$$$).

Your task is to make all symbols in the table equal to $$$0$$$. You are allowed to make at most $$$nm$$$ operations. You don't need to minimize the number of operations.

It can be proved, that it is always possible.

Input format:
The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 5000$$$) — the number of test cases. The next lines contain descriptions of test cases.

The first line of the description of each test case contains two integers $$$n$$$, $$$m$$$ ($$$2 \leq n, m \leq 100$$$).

Each of the next $$$n$$$ lines contains a binary string of length $$$m$$$, describing the symbols of the next row of the table.

It is guaranteed, that the sum of $$$nm$$$ for all test cases does not exceed $$$20000$$$.

Output format:
For each test case print the integer $$$k$$$ ($$$0 \leq k \leq nm$$$) — the number of operations.

In the each of the next $$$k$$$ lines print $$$6$$$ integers $$$x_1, y_1, x_2, y_2, x_3, y_3$$$ ($$$1 \leq x_1, x_2, x_3 \leq n, 1 \leq y_1, y_2, y_3 \leq m$$$) describing the next operation. This operation will be made with three cells $$$(x_1, y_1)$$$, $$$(x_2, y_2)$$$, $$$(x_3, y_3)$$$. These three cells should be different. These three cells should belong to some $$$2 \times 2$$$ square.

Constraints:
Time limit: 1.0 s
Memory limit: 256.0 MB

Note:
In the first test case, it is possible to make only one operation with cells $$$(1, 1)$$$, $$$(2, 1)$$$, $$$(2, 2)$$$. After that, all symbols will be equal to $$$0$$$.

In the second test case:

- operation with cells $$$(2, 1)$$$, $$$(3, 1)$$$, $$$(3, 2)$$$. After it the table will be: 011001000
- operation with cells $$$(1, 2)$$$, $$$(1, 3)$$$, $$$(2, 3)$$$. After it the table will be: 000000000

In the fifth test case:

- operation with cells $$$(1, 3)$$$, $$$(2, 2)$$$, $$$(2, 3)$$$. After it the table will be: 010110
- operation with cells $$$(1, 2)$$$, $$$(2, 1)$$$, $$$(2, 2)$$$. After it the table will be: 000000

Example input:
5
2 2
10
11
3 3
011
101
110
4 4
1111
0110
0110
1111
5 5
01011
11001
00010
11011
10000
2 3
011
101

Example output:
1
1 1 2 1 2 2
2 
2 1 3 1 3 2
1 2 1 3 2 3
4
1 1 1 2 2 2 
1 3 1 4 2 3
3 2 4 1 4 2
3 3 4 3 4 4
4
1 2 2 1 2 2 
1 4 1 5 2 5 
4 1 4 2 5 1
4 4 4 5 3 4
2
1 3 2 2 2 3
1 2 2 1 2 2

Editorial:
Consider two cases: If $$$n = 2$$$ and $$$m = 2$$$, there are only $$$4$$$ possible operations, and we can use up to $$$4$$$ operations. So, one can check all the $$$2^4$$$ possible ways of choosing these operations, and seeing which combination of these operation will result in a full $$$0$$$ grid.
Otherwise, at least one of $$$n$$$ and $$$m$$$ is bigger than $$$2$$$. Without loss of generality imagine $$$n > 2$$$. Take the $$$n$$$th row. For each cell within that row, we can use one operation on it, its left neighbour and the two cells above to fix this cell. We can do this for the first $$$n - 2$$$ cells in the row, and fix the last two with one operation on them. We will make at most $$$n - 1$$$ operations and reach a situation with one empty row. We can take the last row away and apply this procedure for the remaining $$$(n - 1) \times m$$$ grid. If we say inductively that we will have at most $$$(n - 1)m$$$ operations for the remaining grid, we will have done at most $$$(n - 1)m + n - 1 = nm - 1 < nm$$$ operations in total. When $$$n = 2$$$, we can do the same with the columns, and when $$$n = m = 2$$$, we can fix the remaining $$$2 \times 2$$$ grid as we discussed above.
Time complexity: $$$O(nm)$$$ for each case.