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 $p_i \ne p'_i$.


-----Input-----

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.


-----Output-----

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.


-----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, $F(p)=\mathrm{sort}([1+2])=[3]$.

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 $F(p')=\mathrm{sort}([3+1,1+5,5+2,2+4])=\mathrm{sort}([4,6,7,6])=[4,6,6,7]$.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = '\n\ndef main():\n    n = int(input())\n    p = list(map(int,input().split()))\n    p = list(reversed(p))\n    for i in p:\n        print(i, end=" ")\n    print()\n\n\ndef __starting_point():\n    t = int(input())\n    for i in range(t):\n        main()\n"""\n60, 61\n"""\n"""\n"""\n\n__starting_point()'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13339