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].
from sys import stdin, exit
input = stdin.readline

def i(): return input()
def ii(): return int(input())
def iis(): return map(int, input().split())
def liis():	return list(map(int, input().split()))
def print_array(a): print(" ".join(map(str, a)))

def calcular(a):
	soma = 0
	for idx, i in enumerate(a[:-1]):
		soma += abs(a[idx]-a[idx+1])
		
	return soma


t = ii()
for _ in range(t):
	n = ii()	
	p = liis()
	a = p[0]
	numbers1 = [a]
	numbers2 = [a]
	last = p[0]
	cresce = True
	for i in p[1:]:
		if cresce:
			if i >= last:
				last = i
			else:
				numbers1.append(last)
				last = i
				cresce = False
		else:
			if i <= last:
				last = i
			elif numbers1[-1] != last:
				#print(numbers1, last, i)
				numbers1.append(last)
				last = i
				cresce = True
	
	cresce = False
	last = p[0]
	for i in p[1:]:
		if cresce:
			if i >= last:
				last = i
			else:
				numbers2.append(last)
				last = i
				cresce = False
		else:
			if i <= last:
				last = i
			elif last != numbers2[-1]:
				numbers2.append(last)
				last = i
				cresce = True

	if numbers1[-1] != p[-1]: numbers1.append(p[-1])
	if numbers2[-1] != p[-1]: numbers2.append(p[-1])

	#print('n1', numbers1)
	#print('n2', numbers2)


	if calcular(numbers1) > calcular(numbers2):
		print(len(numbers1))
		print_array(numbers1)
	elif calcular(numbers1) == calcular(numbers2):
		if len(numbers1) < len(numbers2):
			print(len(numbers1))
			print_array(numbers1)
		else:
			print(len(numbers2))
			print_array(numbers2)
	else:
		print(len(numbers2))
		print_array(numbers2)

