Equalize by Divide
Contest: Codeforces Round 854 by cybercats (Div. 1 + Div. 2)
Difficulty rating: 1200
Tags: brute force, constructive algorithms, greedy, math

You are given an array $$$a_1, a_2, \ldots, a_n$$$ of positive integers.

You can make this operation multiple (possibly zero) times:

- Choose two indices $$$i$$$, $$$j$$$ ($$$1 \leq i, j \leq n$$$, $$$i \neq j$$$).
- Assign $$$a_i := \lceil \frac{a_i}{a_j} \rceil$$$. Here $$$\lceil x \rceil$$$ denotes $$$x$$$ rounded up to the smallest integer $$$\geq x$$$.

Is it possible to make all array elements equal by some sequence of operations (possibly empty)? If yes, print any way to do it in at most $$$30n$$$ operations.

It can be proven, that under the problem constraints, if some way exists to make all elements equal, there exists a way with at most $$$30n$$$ operations.

Input format:
The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. Descriptions of test cases follow.

The first line of each test case description contains a single integer $$$n$$$ ($$$1 \leq n \leq 100$$$).

The second line of each test case description contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 10^9$$$).

It is guaranteed, that the sum of $$$n$$$ for all test cases does not exceed $$$1000$$$.

Output format:
For each test case print a single integer $$$q$$$ ($$$-1 \leq q \leq 30n$$$). If $$$q=-1$$$, there is no solution, otherwise $$$q$$$ is equal to the number of operations.

If $$$q \geq 0$$$, on the next $$$q$$$ lines print two integers $$$i$$$, $$$j$$$ ($$$1 \leq i, j \leq n$$$, $$$i \neq j$$$) — descriptions of operations.

If there are multiple solutions, you can print any.

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

Note:
In the first and second, fourth test cases all numbers are equal, so it is possible to do nothing.

In the third test case, it is impossible to make all numbers equal.

In the fifth test case: $$$[\color{red}{4}, 3, \color{blue}{2}] \to [\color{blue}{2}, \color{red}{3}, 2] \to [2, 2, 2]$$$.

In the sixth test case: $$$[\color{blue}{3}, 3, \color{red}{4}, 4] \to [3, \color{blue}{3}, 2, \color{red}{4}] \to [\color{red}{3}, 3, \color{blue}{2}, 2] \to [2, \color{red}{3}, 2, \color{blue}{2}] \to [2, 2, 2, 2]$$$.

Here the red numbers are $$$i$$$ indices (that will be assigned), blue numbers are $$$j$$$ indices.

Example input:
10
1
100
3
1 1 1
2
2 1
2
5 5
3
4 3 2
4
3 3 4 4
2
2 100
5
5 3 6 7 8
6
3 3 80 3 8 3
4
19 40 19 55

Example output:
0
0
-1
0
2
1 3
2 1
4
3 1
4 2
1 3
2 4
6
2 1
2 1
2 1
2 1
2 1
2 1
8
5 2
4 2
3 2
1 3
1 3
2 1
4 1
5 1
4
5 1
3 1
3 1
3 1
9
4 2
2 1
1 2
1 2
3 2
3 2
1 4
2 4
3 4

Editorial:
If all numbers are equal initially — we can do nothing.
Otherwise if some $$$a_i = 1$$$, answer do not exist: this $$$a_i$$$ can't became bigger during operations and all other elements can't be equal to $$$1$$$ simultaniously, because after the last operation $$$a_j > 1$$$ (otherwise we can remove this operation).
If all $$$a_i \geq 2$$$, the answer exists and we can simulate such algorithm: let's take $$$i$$$, such that $$$a_i$$$ is maximum possible and $$$j$$$, such that $$$a_j$$$ is smallest possible. Make operation with $$$(i, j)$$$. Note, that after at most $$$30n$$$ operations all elements will be equal.
It is true, because after each operation $$$a_i$$$ decreases at least by $$$2$$$ times (and rounded up) and all elements are bounded $$$a_x \geq 2$$$ after each operation. Each number can't be decreased more than $$$30$$$ times.
Time complexity: $$$O(n^2 \log{C})$$$, where $$$C = 10^9$$$.