1991D · Prime XOR Coloring

Problem: You are given an undirected graph with $n$ vertices, numbered from $1$ to $n$. There is an edge between vertices $u$ and $v$ if and only if $u \oplus v$ is a prime number, where $\oplus$ denotes the bitwise XOR operator.

Color all vertices of the graph using the minimum number of colors, such that no two vertices directly connected by an edge have the same color.

Input Format: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of test cases follows.

The only line contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of vertices in the graph.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output Format: For each test case, output two lines.

The first line should contain a single integer $k$ ($1 \le k \le n$) — the minimum number of colors required.

The second line should contain $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \le c_i \le k$) — the color of each vertex.

If there are multiple solutions, output any of them.

Note: In the first test case, the minimum number of colors is $1$, because there is only one vertex.

In the second test case, the minimum number of colors is $2$, because there is an edge connecting $1$ and $2$ ($1 \oplus 2 = 3$, which is a prime number).

In the third test case, the minimum number of colors is still $2$, because $2$ and $3$ can be colored the same since there is no edge between $2$ and $3$ ($2 \oplus 3 = 1$, which is not a prime number).

In the fourth test case, it can be shown that the minimum number of colors is $3$.

In the fifth test case, it can be shown that the minimum number of colors is $3$.

In the sixth test case, it can be shown that the minimum number of colors is $4$.