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 PYTHON):

Solution:
import sys
import math
 
def remove_duplicates(p):
	n = len(p)
	to_remove = [0 for i in range(n)]
	for i in range(n-1):
		if p[i] == p[i+1]:
			to_remove[i] = 1
	return [p[i] for i in range(n) if not to_remove[i]]


t = int(sys.stdin.readline().strip())
for _ in range(t):
	n = int(sys.stdin.readline().strip())
	p = remove_duplicates(list(map(int, sys.stdin.readline().strip().split(' '))))	
	n = len(p)

	i = 0
	while i < n-1:
		j = i+1
		if p[i+1] > p[i]:
			while j < n-1 and p[j+1] > p[j]:
				p[j] = -1
				j += 1
		elif p[i+1] < p[i]:
			while j < n-1 and p[j+1] < p[j]:
				p[j] = -1
				j += 1
		i = j
	
	ans = 0
	p = [pi for pi in p if pi != -1]
	for i in range(len(p) -1):
		ans += abs(p[i] - p[i+1])

	print(len(p))
	print(' '.join([str(pi) for pi in p]))