human
Write me a Python code to find a different permutation p' with the same fingerprint given the following:
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,

$$$F(p)=sort([p_1+p_2,p_2+p_3,…,p_{n-1}+p_n]).$$$

For example, if n=4 and p=[1,4,2,3], then the fingerprint is given by F(p)=sort([1+4,4+2,2+3])=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 ≠ p'_i.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 668). Description of the test cases follows.

The first line of each test case contains a single integer n (2≤ n≤ 100) — the length of the permutation.

The second line of each test case contains n integers p_1,…,p_n (1≤ p_i≤ n). It is guaranteed that p is a permutation.

Output

For each test case, output n integers p'_1,…, p'_n — a permutation such that p'≠ 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.

Example

Input

3
2
1 2
6
2 1 6 5 4 3
5
2 4 3 1 5

Output

2 1
1 2 5 6 3 4
3 1 5 2 4

Note

In the first test case, F(p)=sort([1+2])=[3].

And F(p')=sort([2+1])=[3].

In the second test case, F(p)=sort([2+1,1+6,6+5,5+4,4+3])=sort([3,7,11,9,7])=[3,7,7,9,11].

And F(p')=sort([1+2,2+5,5+6,6+3,3+4])=sort([3,7,11,9,7])=[3,7,7,9,11].

In the third test case, F(p)=sort([2+4,4+3,3+1,1+5])=sort([6,7,4,6])=[4,6,6,7].

And F(p')=sort([3+1,1+5,5+2,2+4])=sort([4,6,7,6])=[4,6,6,7].
Using python3 can you solve the prior task?
assistant
#Here is Python code to find a different permutation p' with the same fingerprint.

import os
import sys
from io import BytesIO, IOBase
# from collections import defaultdict as dd
# from collections import deque as dq
# import itertools as it
# from math import sqrt, log, log2
# from fractions import Fraction

def main():
    t = int(input())
    for _ in range(t):
        n = int(input())
        #x, y = map(int, input().split())
        nums = list(map(int, input().split()))
        print(*nums[::-1])
        
























# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
    newlines = 0
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
 
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
 
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout =...