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:
import sys
import math 
from decimal import *
def I(): return int(sys.stdin.readline())
def IL(): return list(map(int,sys.stdin.readline().strip().split()))
def ILS(): return list(map(str,sys.stdin.readline().strip().split()))

tt = I()
for ii in range(tt):
	n = I()
	l =IL()
	ans = []
	ans.append(l[0])
	last = l[1]
	trend = ans[-1]-last

	for i in range(2,len(l)) : 
		x = l[i]
		# print("T: ",last,x,last-x,trend)
		if last-x < 0 :
			if trend > 0 : 
				ans.append(last) 
			trend = last-x
			last = x

		else : 
			if trend < 0 : 
				ans.append(last) 
			trend = last-x
			last = x
	ans.append(last)
	print(len(ans))
	for aa in ans : print(aa,end=" ")
	print()
	pass