Make Nonzero Sum (hard version)
Contest: Codeforces Round 829 (Div. 1)
Difficulty rating: 1500
Tags: constructive algorithms, dp, greedy

This is the hard version of the problem. The difference is that in this version the array contains zeros. You can make hacks only if both versions of the problem are solved.

You are given an array $$$[a_1, a_2, \ldots a_n]$$$ consisting of integers $$$-1$$$, $$$0$$$ and $$$1$$$. You have to build a partition of this array into the set of segments $$$[l_1, r_1], [l_2, r_2], \ldots, [l_k, r_k]$$$ with the following property:

- Denote the alternating sum of all elements of the $$$i$$$-th segment as $$$s_i$$$: $$$s_i$$$ = $$$a_{l_i} - a_{l_i+1} + a_{l_i+2} - a_{l_i+3} + \ldots \pm a_{r_i}$$$. For example, the alternating sum of elements of segment $$$[2, 4]$$$ in array $$$[1, 0, -1, 1, 1]$$$ equals to $$$0 - (-1) + 1 = 2$$$.
- The sum of $$$s_i$$$ over all segments of partition should be equal to zero.

Note that each $$$s_i$$$ does not have to be equal to zero, this property is about sum of $$$s_i$$$ over all segments of partition.

The set of segments $$$[l_1, r_1], [l_2, r_2], \ldots, [l_k, r_k]$$$ is called a partition of the array $$$a$$$ of length $$$n$$$ if $$$1 = l_1 \le r_1, l_2 \le r_2, \ldots, l_k \le r_k = n$$$ and $$$r_i + 1 = l_{i+1}$$$ for all $$$i = 1, 2, \ldots k-1$$$. In other words, each element of the array must belong to exactly one segment.

You have to build a partition of the given array with properties described above or determine that such partition does not exist.

Note that it is not required to minimize the number of segments in the partition.

Input format:
Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows.

The first line of each test case contains an integer $$$n$$$ ($$$1 \le n \le 200\,000$$$) — the length of array $$$a$$$.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$a_i$$$ is $$$-1$$$, $$$0$$$, or $$$1$$$) — the elements of the given array.

It's guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$200\,000$$$.

Output format:
For each test case print an integer $$$k$$$ — the number of segments in the partition. If required partition does not exist, print $$$-1$$$.

If partition exists, in the $$$i$$$-th of the following $$$k$$$ lines print two integers $$$l_i$$$ and $$$r_i$$$ — description of the $$$i$$$-th segment. The following conditions should be satisfied:

- $$$l_i \le r_i$$$ for each $$$i$$$ from $$$1$$$ to $$$k$$$.
- $$$l_{i + 1} = r_i + 1$$$ for each $$$i$$$ from $$$1$$$ to $$$(k - 1)$$$.
- $$$l_1 = 1, r_k = n$$$.

If there are multiple correct partitions of the array, print any of them.

Constraints:
Time limit: 2.0 s
Memory limit: 256.0 MB

Note:
In the first test case we can build a partition of $$$4$$$ segments — each of them will contain only one element of the array equals to $$$0$$$. So the sum will be equal to $$$0 + 0 + 0 + 0 = 0$$$.

In the second test case we can build a partition of $$$4$$$ segments. The alternating sum of the first segment will be equal to $$$-1$$$, the alternating sum of the second segment will be equal to $$$1$$$, of the third segment — $$$0 - 1 + 0 = -1$$$, of the fourth segment — $$$1 - 0 = 1$$$. The sum will be equal to $$$-1 + 1 -1 + 1 = 0$$$.

In the third test case it can be proved that the required partition does not exist.

Example input:
5
4
0 0 0 0
7
-1 1 0 1 0 1 0
5
0 -1 1 0 1
3
1 0 1
1
1

Example output:
4
1 1
2 2
3 3
4 4
4
1 1
2 2
3 5
6 7
-1
2
1 1
2 3
-1

Editorial:
If the sum of all numbers in the array is odd, then splitting is impossible, because splitting does not affect the evenness of the sum. Otherwise, we will build the answer constructively. Suppose we have considered some kind of array prefix. Let's keep going until we get exactly $$$2$$$ non-zero numbers.
We want to make these two non-zero numbers add up to $$$0$$$. Then if on the last segment the sum is already equal to $$$0$$$, then just take it as an answer. Otherwise, consider a few cases:
• If the length of the segment is even, then we simply separate the last number (it will be non-zero) into a separate segment. Then its sign will change and in total these two numbers will give $$$0$$$.
• The same can be done if the length of the segment is odd, but its first element is equal to $$$0$$$. Separate this $$$0$$$ and repeat the algorithm above.
• If the length of the segment is odd and the first element is not equal to $$$0$$$, then we separate it. Then the value of the first element will not change, and the last will change to the opposite, and then their sum will be equal to $$$0$$$.