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:
from sys import stdin, stdout
import math
from itertools import permutations, combinations
from collections import defaultdict
from collections import Counter
from bisect import bisect_left
import sys
from queue import PriorityQueue

mod = 1000000007

def L():
	return list(map(int, stdin.readline().split()))

def In():
	return map(int, stdin.readline().split())

def I():
	return int(stdin.readline())

def printIn(ob):
	return stdout.write(str(ob)+'\n')

def powerLL(n, p):
	result = 1
	while (p):
		if (p&1):
			result = result * n % mod
		p = int(p / 2)
		n = n * n % mod
	return result

#--------------------------------------

def myCode():
    n = I()
    lst = L()
    ans_list = [lst[0]]
    for i in range(1,len(lst)-1):
        if lst[i-1]<lst[i]>lst[i+1] or lst[i-1]>lst[i]<lst[i+1]:
            ans_list.append(lst[i])

    ans_list.append(lst[-1])
    print(len(ans_list))
    print(*ans_list)

def main():
    for t in range(I()):
	    myCode()
if __name__ == '__main__':
    # print(math.sqrt(4))
    main()