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A bitwise automaton is a directed acyclic graph with its vertices called states, which has two terminal states, called 0 and 1, and an arbitrary amount of other states numbered with consecutive integers starting from 2. The terminal states have no outgoing edges. Each non-terminal state has exactly two outgoing edges, one marked with 0 and one marked with 1. Additionally, each non-terminal state i has a bit number bi associated with it, and one of the states (either terminal or non-terminal) is denoted as the starting state.
A bitwise automaton can be evaluated on an integer input x in the following way: first, we put a token to the starting state. While the token is in a non-terminal state i, we do the following: check the bi-th bit of x, and move the token to a new state using the outgoing edge with the corresponding label. The bits are numbered from 0 from lowest to highest: the 0-th bit is the parity bit of x, and so on. Eventually the token will reach a terminal state, and the number of this state (either 0 or 1) is the output of the evaluation for the given x.
You are given the desired result of evaluation of an automaton for all inputs from 0 to n−1, and need to construct an automaton with the smallest number of states that gives such evaluation results.
The input contains multiple test cases. The first line of the input contains the number t of test cases, 1 ≤ t ≤ 510. Each test case is then described with two lines. The first of those lines contains one integer n, 1 ≤ n ≤ 8. The second of those lines contains n integers, each either 0 or 1 — the desired results of evaluation for inputs 0, 1, . . . , n − 1.
Print the outputs for all test cases in order. For each test case, first print the number m (m ≥ 2) of states in your automaton on a line of its own. It must be the minimum possible number of states for an automaton that produces the desired results.
In the next m − 2 lines describe the states from 2 to m − 1 in order. Describe a state with three integers: the bit number bi (0 ≤ bi ≤ 2), the number of the state where the 0 edge leads, and the number of the state where the 1 edge leads.
In the next line print the number of the starting state.
In case there are multiple possible solutions, output any.
1 5 1 0 0 0 1
4 0 3 0 1 1 0 2