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:
#!/usr/bin/python3

# Codeforces - Round 649
# Author: frostD
# Problem B - Most Socially Distanced Subsequence


def read_int():
	n = int(input())
	return n


def read_ints():
	ints = [int(x) for x in input().split(" ")]
	return ints
	

#---

def solve(n, p):
	res = n
	inc = True if p[1] > p[0] else False
	idx = [0 for _ in range(n)]
	idx[0] = 1
	idx[-1] = 1
	
	for i in range(1, n-1):
		if inc:
			if p[i+1] > p[i]:
				res -= 1
			else:
				inc = False
				idx[i] = 1
		else:
			if p[i+1] < p[i]:
				res -= 1
			else:
				inc = True
				idx[i] = 1
	
	arr = []
	for i in range(n):
		if idx[i] == 1:
			arr.append(str(p[i]))
			
	return res, arr
	
	

# Main
t = read_int()
for case in range(t):
	n = read_int()
	p = read_ints()
	
	res, arr = solve(n, p)
	print (res)
	print (' '.join(arr))