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 math import *
from collections import *
from operator import itemgetter
import bisect
from heapq import *
i = lambda: input()
ii = lambda: int(input())
iia = lambda: list(map(int,input().split()))
isa = lambda: list(input().split())
I = lambda:list(map(int,input().split()))
chrIdx = lambda x: ord(x)-96
idxChr = lambda x: chr(96+x)

def find(a,i):
	x = bin(i)[2:]
	if(x.count('1')<2):
		return [0,[]]
	else:
		prev = 0
		cur = 0
		new = []
		for i in range(len(x)):
			if(x[i]=='1'):
				new.append(a[i])
		for i in range(len(new)-1):
			cur	+= abs(new[i]-new[i+1])
		return [cur,new]
t = ii()

for _ in range(t):
	n = ii()
	a = iia()
	prev = 0
	maxi = 0
	ans = 0
	"""
	for i in range(1,2**n):
		d = find(a,i)
		#print(d)
		maxi = max(maxi,d[0])
		if(maxi!=prev):
			ans = d[1]
		prev = maxi
	"""
	inc = 0
	ans = []
	i = 0
	while(i<n-1):
		if(i<n-1 and a[i]<a[i+1]):
			if(len(ans)>0 and ans[-1]!=a[i]):
				ans.append(a[i])
			elif(len(ans)==0):
				ans.append(a[i])
			j = i+1
			while(j<n-1 and a[j]<a[j+1]):
				j+=1
			ans.append(a[j])
			i = j
		elif(i<n-1 and a[i]>a[i+1]):
			if(len(ans)>0 and ans[-1]!=a[i]):
				ans.append(a[i])
			elif(len(ans)==0):
				ans.append(a[i])
			j = i+1
			while(j<n-1 and a[j]>a[j+1]):
				j+=1
			ans.append(a[j])
			i = j



	print(len(ans)) 
	print(*ans)
				
