Problem: 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].

Solution (in PYTHON3):

Solution:

import sys
sys.setrecursionlimit(10000)
# default is 1000 in python


t = int(input())
# t = 1

for _ in range(t):
	n = int(input())

	# u, v = list(map(int, input().split()))

	p = list(map(int, input().split()))

	prev = p[1]
	major = [p[0], p[1]]

	ismin = True
	if p[1] > p[0]:
		ismin = False

	# appended = {}

	for no in p[2:]:
		if no < prev and ismin is True:
			# appended[ind] = True
			del major[-1]
			major.append(no)
		elif no > prev and ismin is True:
			major.append(no)
			ismin = False
		elif no < prev and ismin is False:
			major.append(no)
			ismin = True
		elif no > prev and ismin is False:
			del major[-1]
			major.append(no)
		prev = no
	print(len(major))
	print(*major)


# try:
	# raise Exception
# except:
	# print("-1")
	

# thenos.sort(key=lambda x: x[2], reverse=True)

# int(math.log(max(numbers)+1,2))

# 2**3 (power)

# a,t = (list(x) for x in zip(*sorted(zip(a, t))))

# to copy lists use .copy()

# pow(p, si, 1000000007) for modular exponentiation

# my_dict.pop('key', None)
# This will return my_dict[key] if key exists in the dictionary, and None otherwise.