A. Circle Coloringtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given three sequences: a_1, a_2, \ldots, a_n; b_1, b_2, \ldots, b_n; c_1, c_2, \ldots, c_n.For each i, a_i \neq b_i, a_i \neq c_i, b_i \neq c_i.Find a sequence p_1, p_2, \ldots, p_n, that satisfy the following conditions: p_i \in \{a_i, b_i, c_i\} p_i \neq p_{(i \mod n) + 1}.In other words, for each element, you need to choose one of the three possible values, such that no two adjacent elements (where we consider elements i,i+1 adjacent for i<n and also elements 1 and n) will have equal value.It can be proved that in the given constraints solution always exists. You don't need to minimize/maximize anything, you need to find any proper sequence.InputThe first line of input contains one integer t (1 \leq t \leq 100): the number of test cases.The first line of each test case contains one integer n (3 \leq n \leq 100): the number of elements in the given sequences.The second line contains n integers a_1, a_2, \ldots, a_n (1 \leq a_i \leq 100).The third line contains n integers b_1, b_2, \ldots, b_n (1 \leq b_i \leq 100).The fourth line contains n integers c_1, c_2, \ldots, c_n (1 \leq c_i \leq 100).It is guaranteed that a_i \neq b_i, a_i \neq c_i, b_i \neq c_i for all i.OutputFor each test case, print n integers: p_1, p_2, \ldots, p_n (p_i \in \{a_i, b_i, c_i\}, p_i \neq p_{i \mod n + 1}).If there are several solutions, you can print any.ExampleInput
5
3
1 1 1
2 2 2
3 3 3
4
1 2 1 2
2 1 2 1
3 4 3 4
7
1 3 3 1 1 1 1
2 4 4 3 2 2 4
4 2 2 2 4 4 2
3
1 2 1
2 3 3
3 1 2
10
1 1 1 2 2 2 3 3 3 1
2 2 2 3 3 3 1 1 1 2
3 3 3 1 1 1 2 2 2 3
Output
1 2 3
1 2 1 2
1 3 4 3 2 4 2
1 3 2
1 2 3 1 2 3 1 2 3 2
NoteIn the first test case p = [1, 2, 3].It is a correct answer, because:  p_1 = 1 = a_1, p_2 = 2 = b_2, p_3 = 3 = c_3  p_1 \neq p_2 , p_2 \neq p_3 , p_3 \neq p_1 All possible correct answers to this test case are: [1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1].In the second test case p = [1, 2, 1, 2].In this sequence p_1 = a_1, p_2 = a_2, p_3 = a_3, p_4 = a_4. Also we can see, that no two adjacent elements of the sequence are equal.In the third test case p = [1, 3, 4, 3, 2, 4, 2].In this sequence p_1 = a_1, p_2 = a_2, p_3 = b_3, p_4 = b_4, p_5 = b_5, p_6 = c_6, p_7 = c_7. Also we can see, that no two adjacent elements of the sequence are equal.
5
3
1 1 1
2 2 2
3 3 3
4
1 2 1 2
2 1 2 1
3 4 3 4
7
1 3 3 1 1 1 1
2 4 4 3 2 2 4
4 2 2 2 4 4 2
3
1 2 1
2 3 3
3 1 2
10
1 1 1 2 2 2 3 3 3 1
2 2 2 3 3 3 1 1 1 2
3 3 3 1 1 1 2 2 2 3

1 2 3 1 2 1 2 1 3 4 3 2 4 2 1 3 2 1 2 3 1 2 3 1 2 3 2 
['constructive algorithms', '*800']