Problem Statement

There are $2N$ gems arranged in a row from left to right, and the type of the $i$-th gem from the left is $A_i$. Here, $A_i$ is an integer between $1$ and $N$, inclusive, and there are exactly two gems of each type.

You want to insert dividers satisfying the following conditions.

• Each divider's position is represented by an integer $i$ ($1 \le i \le 2N - 1$), meaning it divides between the $i$-th and $(i+1)$-th gems from the left.
• No two or more dividers exist at the same position.
• The number of dividers $K$ is at most $N$.
• Inserting $K$ dividers divides the row of gems into $K + 1$ segments. Here, collecting all gems in the odd-numbered (first, third, ...) segments from the left yields exactly one gem of each type $1, 2, \ldots, N$.

Output one way to insert dividers satisfying this condition. It can be proved that such a way always exists.

You are given $T$ test cases; solve each one.

Constraints

• $1 \le T \le 10^5$
• $1 \le N \le 2 \times 10^5$
• $1 \le A_i \le N$ ($1 \le i \leq 2N$)
• There are exactly two gems of each type. That is, for each $x$ ($1 \le x \le N$), there are exactly two indices $i$ with $A_i = x$.
• The sum of $N$ over all test cases is at most $2 \times 10^5$.
• All input values are integers.

Input

The input is given from Standard Input in the following format:

T
\text{case}_1
\text{case}_2
\vdots
\text{case}_T

Each test case is given in the following format:

N
A_1 A_2 \ldots A_{2N}

Output

Output for the $T$ test cases in order.

For each test case, output a way to insert dividers satisfying the condition. Specifically, let $K$ be the number of dividers and $C_1, C_2, \ldots, C_K$ be the positions of the dividers in ascending order, and output in the following format:

K
C_1 C_2 \ldots C_K

Here, $K$ must be at most $N$.