Bakry and Partitioning
Contest: Codeforces Round 746 (Div. 2)
Difficulty rating: 1700
Tags: bitmasks, constructive algorithms, dfs and similar, dp, graphs, trees

Bakry faced a problem, but since he's lazy to solve it, he asks for your help.

You are given a tree of $$$n$$$ nodes, the $$$i$$$-th node has value $$$a_i$$$ assigned to it for each $$$i$$$ from $$$1$$$ to $$$n$$$. As a reminder, a tree on $$$n$$$ nodes is a connected graph with $$$n-1$$$ edges.

You want to delete at least $$$1$$$, but at most $$$k-1$$$ edges from the tree, so that the following condition would hold:

- For every connected component calculate the bitwise XOR of the values of the nodes in it. Then, these values have to be the same for all connected components.

Is it possible to achieve this condition?

Input format:
Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ $$$(1 \leq t \leq 5 \cdot 10^4)$$$. Description of the test cases follows.

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ $$$(2 \leq k \leq n \leq 10^5)$$$.

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

The $$$i$$$-th of the next $$$n-1$$$ lines contains two integers $$$u_i$$$ and $$$v_i$$$ ($$$1 \leq u_i, v_i \leq n$$$, $$$u_i\neq v_i$$$), which means that there's an edge between nodes $$$u_i$$$ and $$$v_i$$$.

It is guaranteed that the given graph is a tree.

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

Output format:
For each test case, you should output a single string. If you can delete the edges according to the conditions written above, output "YES" (without quotes). Otherwise, output "NO" (without quotes).

You can print each letter of "YES" and "NO" in any case (upper or lower).

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

Note:
It can be shown that the objection is not achievable for first, third, and fifth test cases.

In the second test case, you can just remove all the edges. There will be $$$5$$$ connected components, each containing only one node with value $$$3$$$, so the bitwise XORs will be $$$3$$$ for all of them.

In the fourth test case, this is the tree: .

You can remove an edge $$$(4,5)$$$

The bitwise XOR of the first component will be, $$$a_1 \oplus a_2 \oplus a_3 \oplus a_4 = 1 \oplus 6 \oplus 4 \oplus 1 = 2$$$ (where $$$\oplus$$$ denotes the bitwise XOR).

The bitwise XOR of the second component will be, $$$a_5 = 2$$$.

Example input:
5
2 2
1 3
1 2
5 5
3 3 3 3 3
1 2
2 3
1 4
4 5
5 2
1 7 2 3 5
1 2
2 3
1 4
4 5
5 3
1 6 4 1 2
1 2
2 3
1 4
4 5
3 3
1 7 4
1 2
2 3

Example output:
NO
YES
NO
YES
NO

Editorial:
The most important observation is: If you can partition the tree into $$$m$$$ components such that the xor of every component is $$$x$$$, Then you can partition the tree into $$$m-2$$$ components by merging any 3 adjacent components into 1 component, and the xor of the new component will equal $$$x$$$, since $$$x$$$ xor $$$x$$$ xor $$$x$$$ = $$$x$$$.
Notice that the answer is always YES if the xor of the array is $$$0$$$. Because you can delete any edge in the tree, and the 2 components will have the same xor.
Otherwise, We need to partition the tree into 3 components that have the same xor. Let $$$x$$$ be the xor of all node values in the tree, then The xor of every component will equal $$$x$$$.
We need to search for 2 edges to delete from the tree and one of them such that the xor every component equals $$$x$$$ and if we found them and $$$K \neq 2$$$ then the answer is "YES" otherwise "NO".
To search on the 2 edges, We will first root the tree on node 1, then we will search on the deepest subtree such that xor value of subtree equals $$$x$$$, then erase the edge and search again for the 2nd edge.