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 collections import Counter as C,defaultdict as D,deque as Q
from operator import itemgetter as I
from itertools import product as P
from bisect import bisect_left as BL,bisect_right as BR
from heapq import heappush as HPUSH,heappop as HPOP
from math import floor as MF,ceil as MC, gcd as MG,factorial as F,sqrt as SQRT, inf as INFINITY


def player1():
	print("")

def player2():
	print("")

def isPrime(n):

	for i in range(2,MC(SQRT(n))+1):
		if n%i==0:
			return False
	return True


def main():


	for _ in range(int(input())):

		n=int(input())

		p=list(map(int,input().split( )))
		ans=[]
		ans.append(p[0])
		if p[1]>p[0]:
			ans.append(p[1])
		else:
			ans.append(-p[1])

		for i in range(2,n):

			if p[i]>abs(ans[-1]):
				ans.append(-abs(ans[-1]))
				ans.append(p[i])
			else:
				ans.append(abs(ans[-1]))
				ans.append(-p[i])
		
		res=[ans[0]]
		for i in range(1,len(ans)-1,2):
			if ans[i]!=-ans[i+1]:
				res.append(abs(ans[i]))

		res.append(abs(ans[-1]))
		print(len(res))
		print(*res)






main()
