human
You have a permutation: an array a = [a_1, a_2, …, a_n] of distinct integers from 1 to n. The length of the permutation n is odd.

You need to sort the permutation in increasing order.

In one step, you can choose any prefix of the permutation with an odd length and reverse it. Formally, if a = [a_1, a_2, …, a_n], you can choose any odd integer p between 1 and n, inclusive, and set a to [a_p, a_{p-1}, …, a_1, a_{p+1}, a_{p+2}, …, a_n].

Find a way to sort a using no more than 5n/2 reversals of the above kind, or determine that such a way doesn't exist. The number of reversals doesn't have to be minimized.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.

The first line of each test case contains a single integer n (3 ≤ n ≤ 2021.
n is odd) — the length of the permutation.

The second line contains n distinct integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n) — the permutation itself. 

It is guaranteed that the sum of n over all test cases does not exceed 2021.

Output

For each test case, if it's impossible to sort the given permutation in at most 5n/2 reversals, print a single integer -1.

Otherwise, print an integer m (0 ≤ m ≤ 5n/2), denoting the number of reversals in your sequence of steps, followed by m integers p_i (1 ≤ p_i ≤ n.
p_i is odd), denoting the lengths of the prefixes of a to be reversed, in chronological order.

Note that m doesn't have to be minimized. If there are multiple answers, print any.

Example

Input

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

Output

4
3 3 3 3
2
3 5
-1

Note

In the first test case, the permutation is already sorted. Any even number of reversals of the length 3 prefix doesn't change that fact.

In the second test case, after reversing the prefix of length 3 the permutation will change to [5, 4, 3, 2, 1], and then after reversing the prefix of length 5 the permutation will change to [1, 2, 3, 4, 5].

In the third test case, it's impossible to sort the permutation.
Using python3 can you solve the prior task?
assistant
#Here is Python code to find a way to sort a using no more than 5n/2 reversals of the above kind, or determine that such a way doesn't exist.

#!/usr/bin/env python3
import sys
import getpass  # not available on codechef
import math, random
import functools, itertools, collections, heapq, bisect
from collections import Counter, defaultdict, deque
input = sys.stdin.readline  # to read input quickly

# available on Google, AtCoder Python3, not available on Codeforces
# import numpy as np
# import scipy

M9 = 10**9 + 7  # 998244353
yes, no = "YES", "NO"
# d4 = [(1,0),(0,1),(-1,0),(0,-1)]
# d8 = [(1,0),(1,1),(0,1),(-1,1),(-1,0),(-1,-1),(0,-1),(1,-1)]
# d6 = [(2,0),(1,1),(-1,1),(-2,0),(-1,-1),(1,-1)]  # hexagonal layout
MAXINT = sys.maxsize

# if testing locally, print to terminal with a different color
OFFLINE_TEST = getpass.getuser() == "hkmac"
# OFFLINE_TEST = False  # codechef does not allow getpass
def log(*args):
    if OFFLINE_TEST:
        print('\033[36m', *args, '\033[0m', file=sys.stderr)

def solve(*args):
    # screen input
    if OFFLINE_TEST:
        log("----- solving ------")
        log(*args)
        log("----- ------- ------")
    return solve_(*args)

def read_matrix(rows):
    return [list(map(int,input().split())) for _ in range(rows)]

def read_strings(rows):
    return [input().strip() for _ in range(rows)]

def minus_one(arr):
    return [x-1 for x in arr]

def minus_one_matrix(mrr):
    return [[x-1 for x in row] for row in mrr]

# ---------------------------- template ends here ----------------------------

def solve_(arr):
    # your solution here
    
    for i,x in enumerate(arr, start=1):
        if i%2 != x%2:
            return []

    # arr = [x-1 for x in arr]

    res = []

    def flip(idx):
        if idx == 1:
            return
        nonlocal res
        nonlocal arr
        res.append(idx)
        arr = arr[:idx][::-1] + arr[idx:]
        # log(idx, arr)
        
    for i in range(len(arr)//2):
        right = len(arr) - i*2
        k = len(arr) - i*2
        log(right)

        # bring k to leftmost
        idx = arr.index(k)
        flip(idx+1)

        # bring k to left of k-1
        idx = arr.index(k-1)
...