CF1878G.wxhtzdy ORO Tree

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题目描述

After (finally) qualifying for the IOI 2023, wxhtzdy was very happy, so he decided to do what most competitive programmers do: trying to guess the problems that will be on IOI. During this process, he accidentally made a problem, which he thought was really cool.

You are given a tree (a connected acyclic graph) with nn vertices and n1n-1 edges. Vertex ii ( 1in1 \le i \le n ) has a value aia_i .

Lets' define g(u,v)g(u, v) as the bitwise or of the values of all vertices on the shortest path from uu to vv . For example, let's say that we want to calculate g(3,4)g(3, 4) , on the tree from the first test case in the example. On the path from 33 to 44 are vertices 33 , 11 , 44 . Then, g(3,4)=a3  a1  a4g(3, 4) = a_3 \ | \ a_1 \ | \ a_4 (here, | represents the bitwise OR operation).

Also, you are given qq queries, and each query looks like this:

You are given xx and yy . Let's consider all vertices zz such that zz is on the shortest path from xx to yy (inclusive).

Lets define the niceness of a vertex zz as the sum of the number of non-zero bits in g(x,z)g(x, z) and the number of non-zero bits in g(y,z)g(y, z) . You need to find the maximum niceness among all vertices zz on the shortest path from xx to yy .

Since his brain is really tired after solving an output only problem on SIO (he had to do it to qualify for the IOI), he wants your help with this problem.

输入格式

The first line of input contains a single integer tt ( 1t1041 \le t \le 10^4 ) — the number of test cases.

The first line of each test case contains a single integer nn ( 1n21051 \le n \le 2 \cdot 10^5 ) — the number of vertices.

The second line of each test case contains nn positive integers a1,a2,,ana_1, a_2, \dots, a_n ( 1av1091 \le a_v \le 10^9 ) — the value of each vertex, the ii -th integer in this line corresponds to the vertex ii .

Following n1n - 1 lines are the description of a tree.

Each line contains two integers uu and vv ( 1u,vn,uv1 \le u, v \le n, u \ne v ) — indicating that vertices uu and vv are connected by an edge.

The next line contains a single integer qq ( 1q1051 \le q \le 10^5 ) — number of queries.

Following qq lines contain 2 integers x,yx, y ( 1x,yn1 \le x, y \le n ) — the vertices xx and yy for each query.

It is guaranteed that the sum of nn over all test cases does not exceed 21052 \cdot 10^5 .

It is guaranteed that the sum of qq over all test cases does not exceed 10510^5 .

输出格式

For each test case output qq integers, each of which is the answer to the corresponding query.

输入输出样例

  • 输入#1

    3
    4
    1 2 3 4
    1 3
    1 2
    1 4
    3
    1 1
    1 3
    1 4
    3
    7 6 3
    3 1
    2 1
    4
    1 1
    1 2
    1 3
    2 3
    1
    4
    1
    1 1

    输出#1

    2 4 3 
    6 6 6 6 
    2
  • 输入#2

    3
    7
    4 7 7 4 10 8 10
    6 1
    3 1
    2 1
    7 4
    1 5
    4 2
    4
    7 5
    2 3
    4 5
    2 5
    6
    9 5 6 2 4 6
    5 1
    2 1
    1 6
    4 3
    1 3
    4
    6 1
    1 4
    4 3
    3 5
    7
    5 1 3 7 5 1 6
    2 1
    5 4
    2 3
    3 4
    7 6
    6 3
    2
    4 2
    7 7

    输出#2

    8 6 7 7 
    6 6 4 7 
    6 4
  • 输入#3

    1
    7
    6 8 7 2 5 8 7
    2 1
    3 2
    4 3
    4 6
    4 5
    6 7
    4
    1 5
    6 7
    4 5
    1 4

    输出#3

    7 7 5 7

说明/提示

The image below shows the tree from the second example, first test case.

Tree from the second example, first test caseIn the first query, we have x=7x=7 , y=5y=5 . The shortest path from 77 to 55 is 742157-4-2-1-5 .

Let's calculate the niceness of vertex 77 on this path. We have g(7,7)=a7=10=(1010)2g(7,7)=a_7=10=(1010)_2 and g(5,7)=a5  a1  a2  a4  a7=10  4  7  4  10=15=(1111)2g(5,7)=a_5 \ | \ a_1 \ | \ a_2 \ | \ a_4 \ | \ a_7=10 \ | \ 4 \ | \ 7 \ | \ 4 \ | \ 10=15=(1111)_2 , so its niceness is equal to 2+4=62 + 4 = 6 .

Now let's calculate the niceness of vertex 44 on this path. We have g(7,4)=a7  a4=10  4=14=(1110)2g(7,4)=a_7 \ | \ a_4=10 \ | \ 4=14=(1110)_2 and g(5,4)=a5  a1  a2  a4=10  4  7  4=15=(1111)2g(5,4)=a_5 \ | \ a_1 \ | \ a_2 \ | \ a_4=10 \ | \ 4 \ | \ 7 \ | \ 4=15=(1111)_2 , so its niceness is equal to 3+4=73 + 4 = 7 .

Now let's calculate the niceness of vertex 22 on this path. We have g(7,2)=a7  a4  a2=10  4  7=15=(1111)2g(7,2)=a_7 \ | \ a_4 \ | \ a_2=10 \ | \ 4 \ | \ 7=15=(1111)_2 and g(5,2)=a5  a1  a2=10  4  7=15=(1111)2g(5,2)=a_5 \ | \ a_1 \ | \ a_2=10 \ | \ 4 \ | \ 7=15=(1111)_2 , so its niceness is equal to 4+4=84 + 4 = 8 .

Now let's calculate the niceness of vertex 11 on this path. We have g(7,1)=a7  a4  a2  a1=10  4  7  4=15=(1111)2g(7,1)=a_7 \ | \ a_4 \ | \ a_2 \ | \ a_1=10 \ | \ 4 \ | \ 7 \ | \ 4=15=(1111)_2 and g(5,1)=a5  a1=10  4=14=(1110)2g(5,1)=a_5 \ | \ a_1=10 \ | \ 4=14=(1110)_2 , so its niceness is equal to 4+3=74 + 3 = 7 .

Finally, let's calculate the niceness of vertex 55 on this path. We have g(7,5)=a7  a4  a2  a1  a5=10  4  7  4  10=15=(1111)2g(7,5)=a_7 \ | \ a_4 \ | \ a_2 \ | \ a_1 \ | \ a_5=10 \ | \ 4 \ | \ 7 \ | \ 4 \ | \ 10=15=(1111)_2 and g(5,5)=a5=10=(1010)2g(5,5)=a_5=10=(1010)_2 , so its niceness is equal to 4+2=64 + 2 = 6 .

The maximum niceness on this path is at vertex 22 , and it is 88 .

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