Permutation Forgery
A permutation of length $$$n$$$ is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array) and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array).Let $$$p$$$ be any permutation of length $$$n$$$. We define the fingerprint $$$F(p)$$$ of $$$p$$$ as the sorted array of sums of adjacent elements in $$$p$$$. More formally,$$$$$$F(p)=\mathrm{sort}([p_1+p_2,p_2+p_3,\ldots,p_{n-1}+p_n]).$$$$$$For example, if $$$n=4$$$ and $$$p=[1,4,2,3],$$$ then the fingerprint is given by $$$F(p)=\mathrm{sort}([1+4,4+2,2+3])=\mathrm{sort}([5,6,5])=[5,5,6]$$$.You are given a permutation $$$p$$$ of length $$$n$$$. Your task is to find a different permutation $$$p'$$$ with the same fingerprint. Two permutations $$$p$$$ and $$$p'$$$ are considered different if there is some index $$$i$$$ such that $$$p_i \ne p'_i$$$.
In the first test case, $$$F(p)=\mathrm{sort}([1+2])=[3]$$$.And $$$F(p')=\mathrm{sort}([2+1])=[3]$$$.In the second test case, $$$F(p)=\mathrm{sort}([2+1,1+6,6+5,5+4,4+3])=\mathrm{sort}([3,7,11,9,7])=[3,7,7,9,11]$$$.And $$$F(p')=\mathrm{sort}([1+2,2+5,5+6,6+3,3+4])=\mathrm{sort}([3,7,11,9,7])=[3,7,7,9,11]$$$.In the third test case, $$$F(p)=\mathrm{sort}([2+4,4+3,3+1,1+5])=\mathrm{sort}([6,7,4,6])=[4,6,6,7]$$$.And $$$F(p')=\mathrm{sort}([3+1,1+5,5+2,2+4])=\mathrm{sort}([4,6,7,6])=[4,6,6,7]$$$.
Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 668$$$). Description of the test cases follows.The first line of each test case contains a single integer $$$n$$$ ($$$2\le n\le 100$$$)  — the length of the permutation.The second line of each test case contains $$$n$$$ integers $$$p_1,\ldots,p_n$$$ ($$$1\le p_i\le n$$$). It is guaranteed that $$$p$$$ is a permutation.
For each test case, output $$$n$$$ integers $$$p'_1,\ldots, p'_n$$$ — a permutation such that $$$p'\ne p$$$ and $$$F(p')=F(p)$$$.We can prove that for every permutation satisfying the input constraints, a solution exists.If there are multiple solutions, you may output any.
def i():
    input()
    s = input().split()
    print(' '.join(reversed(s)))
t = int(input())
for _ in range(t):
    i()
def i():
    n = int(input())
    s = list(input().split())
    return ' '.join(reversed(s))
t = int(input())
a = []
for _ in range(t):
    a.append(i())
print('\n'.join(a))
lines = int(input())
for idx in range(lines):
    temp = input()
    print(' '.join(input().split(' ')[::-1]))