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].
#include <CodeforcesSolutions.h>
#include <ONLINE_JUDGE <solution.cf(contestID = "1364",problemID = "A",method = "GET")>.h>
"""
    Author : thekushalghosh
    Team   : CodeDiggers
 
    I prefer Python language over the C++ language :p :D
        
    Visit my website : thekushalghosh.github.io
"""
import sys,math,cmath,time
start_time = time.time()
##########################################################################
################# ---- THE ACTUAL CODE STARTS BELOW ---- #################

def solve():
    n = inp()
    a = inlt()
    q = [a[0]]
    if a[0] < a[1]:
        qw = 1
    else:
        qw = 0
    if a[-1] < a[-2]:
        a.append(1000000000)
    else:
        a.append(0)
    for i in range(len(a) - 1):
        if qw == 1:
            if a[i + 1] < a[i]:
                q.append(a[i])
                qw = 0
        if qw == 0:
            if a[i + 1] > a[i]:
                q.append(a[i])
                qw = 1
    print(len(q))
    print(*q)
            
################## ---- THE ACTUAL CODE ENDS ABOVE ---- ##################
##########################################################################
 
def main():
    global tt
    if not ONLINE_JUDGE:
        sys.stdin = open("input.txt","r")
        sys.stdout = open("output.txt","w")
    t = 1
    t = inp()
    for tt in range(t):
        solve()
    if not ONLINE_JUDGE:
        print("Time Elapsed :",time.time() - start_time,"seconds")
        sys.stdout.close()
    
#---------------------- USER DEFINED INPUT FUNCTIONS ----------------------#
def inp():
    return(int(input()))
def inlt():
    return(list(map(int,input().split())))
def insr():
    return(input().strip())
def invr():
    return(map(int,input().split()))
 
#------------------ USER DEFINED PROGRAMMING FUNCTIONS ------------------#
def counter(a):
    q = [0] * max(a)
    for i in range(len(a)):
        q[a[i] - 1] = q[a[i] - 1] + 1
    return(q)
 
def string_counter(a):
    q = [0] * 26
    for i in range(len(a)):
        q[ord(a[i]) - 97] = q[ord(a[i]) - 97] + 1
    return(q)

def power_two(n):
    if n == 1:
        return(True)
    sttr = str(n)
    len_str = len(sttr)
    sttr = list(sttr)
    num = 0
    if (len_str == 1 and sttr[len_str - 1] == '1'): 
        return(False)
    while (len_str != 1 or sttr[len_str - 1] != '1'):
        if ((ord(sttr[len_str - 1]) - ord('0')) % 2 == 1):
            return(False)
        j = 0; 
        for i in range(len_str):  
            num = num * 10 + (ord(sttr[i]) - ord('0')); 
            if (num < 2): 
                if (i != 0):  
                    sttr[j] = '0'
                    j += 1
                continue
  
            sttr[j] = chr((num // 2) + ord('0'))
            j += 1; 
            num = (num) - (num // 2) * 2
        len_str = j
    return(True)

def log_two(n): 
    return(math.log(n) / math.log(2))

#-----------------------------------------------------------------------#
        
ONLINE_JUDGE = __debug__
if ONLINE_JUDGE:
    input = sys.stdin.readline
    
main()