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:
#===========Template===============
from io import BytesIO, IOBase
from math import sqrt
import sys,os
inpl=lambda:list(map(int,input().split()))
inpm=lambda:map(int,input().split())
inpi=lambda:int(input())
inp=lambda:input()
rev,ra,l=reversed,range,len

P=print
BUFSIZE = 8192
class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def input(): return sys.stdin.readline().rstrip("\r\n")

def B(n):
    return bin(n).replace("0b","")

def factors(n):
    arr=[]
    for i in ra(2,int(sqrt(n))+1):
        if n%i==0:
            arr.append(i)
            if i*i!=n:
                arr.append(int(n/i))
    return arr  

def dfs(arr,n):
    pairs=[[i+1] for i in ra(n)]
    vis=[0]*(n+1)
    for i in ra(l(arr)):
        pairs[arr[i][0]-1].append(arr[i][1])
    pairs[arr[i][1]-1].append(arr[i][0])
    
    comp=[]
    for i in ra(n):
        stack=[pairs[i]]
        temp=[]
        if vis[stack[-1][0]]==0: 
            while len(stack)>0:
                if vis[stack[-1][0]]==0:
                    temp.append(stack[-1][0])
                    vis[stack[-1][0]]=1
                    s=stack.pop()
                    for j in s[1:]:
                        if vis[j]==0:
                            stack.append(pairs[j-1])
                else:
                    stack.pop()
            comp.append(temp)
    return comp
    
  
#=========I/p O/p ========================================#
from bisect import bisect_left as bl 
from bisect import bisect_right as br
import sys,operator,math,operator
import random
#==============To chaliye shuru krte he ==================#
tc=inpi()
while tc:
    tc-=1
    n=inpi()
    li=inpl()
    ans=[li[0]]
    i=1
    idx=0
    while i<n-1:
        a=abs(li[i-1]-li[i+1])
        b=abs(li[i]-li[i-1])+abs(li[i+1]-li[i])
        if a>=b:
            pass
        else:
            ans.append(li[i])
        i+=1
    if idx!=n-1:
        ans.append(li[-1])
    P(l(ans))
    P(*ans)