Given a permutation p of length n, find its subsequence s_1, s_2, …, s_k of length at least 2 such that:

  * |s_1-s_2|+|s_2-s_3|+…+|s_{k-1}-s_k| is as big as possible over all subsequences of p with length at least 2. 
  * Among all such subsequences, choose the one whose length, k, is as small as possible. 



If multiple subsequences satisfy these conditions, you are allowed to find any of them.

A sequence a is a subsequence of an array b if a can be obtained from b by deleting some (possibly, zero or all) elements.

A permutation of length n is an array of length n in which every element from 1 to n occurs exactly once.

Input

The first line contains an integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer n (2 ≤ n ≤ 10^5) — the length of the permutation p.

The second line of each test case contains n integers p_1, p_2, …, p_{n} (1 ≤ p_i ≤ n, p_i are distinct) — the elements of the permutation p.

The sum of n across the test cases doesn't exceed 10^5.

Output

For each test case, the first line should contain the length of the found subsequence, k. The second line should contain s_1, s_2, …, s_k — its elements.

If multiple subsequences satisfy these conditions, you are allowed to find any of them.

Example

Input


2
3
3 2 1
4
1 3 4 2


Output


2
3 1 
3
1 4 2 

Note

In the first test case, there are 4 subsequences of length at least 2:

  * [3,2] which gives us |3-2|=1. 
  * [3,1] which gives us |3-1|=2. 
  * [2,1] which gives us |2-1|=1. 
  * [3,2,1] which gives us |3-2|+|2-1|=2. 



So the answer is either [3,1] or [3,2,1]. Since we want the subsequence to be as short as possible, the answer is [3,1].
production = True

import sys, math

def input(input_format = 0, multi = 0):

    if multi > 0: return [input(input_format) for i in range(multi)]
    else:
        next_line = sys.stdin.readline()

        if input_format >= 10:
            use_list = False
            input_format = int(str(input_format)[-1])
        else: use_list = True

        if input_format == 0: formatted_input = [next_line]
        elif input_format == 1: formatted_input = list(map(int, next_line.split()))
        elif input_format == 2: formatted_input = list(map(float, next_line.split()))
        elif input_format == 3: formatted_input = next_line.split()
        elif input_format == 4: formatted_input = (lambda x: [x[0], list(map(int, x[1:]))])(next_line.split())
        elif input_format == 5: formatted_input = (lambda x: [x[0], list(map(float, x[1:]))])(next_line.split())
        else: formatted_input = [next_line]

        return formatted_input if use_list else formatted_input[0]


def out(output_line, output_format = 0, newline = True):

    formatted_output = ""

    if output_format == 0: formatted_output = str(output_line)
    elif output_format == 1: formatted_output = " ".join(map(str, output_line))
    elif output_format == 1: formatted_output = "\n".join(map(str, output_line))

    print(formatted_output, end = "\n" if newline else "")


def log(*args):
    if not production:
        print("$$$")
        print(*args)


#
#   >>>>>>>>>>>>>>> START OF SOLUTION <<<<<<<<<<<<<<
#


def solve():

    n = input(1)

    p = input(1)

    a = [p[0]]
    d = -1

    for i in p[1:]:
        if i > a[-1]:
            if d == 0:
                a.pop()
            a.append(i)
            d = 0
        elif i < a[-1]:
            if d == 1:
                a.pop()
            a.append(i)
            d = 1

    if len(a) < 2:
        a.append(p[-1])

    out(len(a))
    out(a, 1)

    return


for i in range(input(11)): solve()
# solve()

#
#   >>>>>>>>>>>>>>>> END OF SOLUTION <<<<<<<<<<<<<<<
#
