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].
# May the Speedforce be with us...
'''
problem statement achhe se padhna hai
age se bhi dekhna hai, piche se bhi dekhna hai
'''
#from math import gcd,ceil,sin,radians,cos
#from bisect import insort
#from collections import defaultdict as dd
def lcm(a,b):
	return (a*b)//gcd(a,b)
def inin():
	return int(input())
def inar():
	return list(map(int,input().split()))
def ar(element,size):
	return [element for i in range(size)]
def yes():
	print('Yes')
def no():
	print('No')
'''
def countDivisibles(A, B, M): #count numbers in the range[A,B] which are divisible by M
    if (A % M == 0): 
        return ((B // M) - (A // M)) + 1
    return ((B // M) - (A // M)) 
	
def  countSetBits(n): #count the number of set bits in the binary representation of an integer
    count = 0
    while (n): 
        count += n & 1
        n >>= 1
    return count
def digitsum(num): #return the sum of dogots of a number
	su=0
	while(num):
		su+=num%10
		num//=10
	return su
'''

Testcase=inin()
for _ in range(Testcase):
	n=inin()
	a=inar()
	curr=a[0]
	i=1
	ans=[curr]
	while(i<n):
		if a[i]>curr:
			while(i<n and a[i]>curr):
				curr=a[i]
				i+=1
			ans.append(curr)
			continue

		if a[i]<curr:
			while(i<n and a[i]<curr):
				curr=a[i]
				i+=1
			ans.append(curr)
			continue
	print(len(ans))
	print(*ans)

