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].
"""
~~ Author : Bhaskar
~~ Dated : 07~06~2020
"""

import sys
from bisect import *
from math import floor,sqrt,ceil,factorial as F,gcd,pi,log
from itertools import chain,combinations,permutations,accumulate
from collections import Counter,defaultdict,OrderedDict,deque
from functools import cmp_to_key,reduce
INT_MAX = sys.maxsize
INT_MIN = -(sys.maxsize)-1
mod = 1000000007
ch = "abcdefghijklmnopqrstuvwxyz"
lcm = lambda a,b : (a*b)//gcd(a,b)
setbit = lambda x : bin(x)[2:].count("1")

def isPick(a,n,x,i,j):
    if i >= 0 and a[i] > x:
        return False
    if j < n and a[j] > x:
        return False
    return True

def isDown(a,n,x,i,j):
    if i >= 0 and a[i] < x:
        return False
    if j < n and a[j] < x:
        return False
    return True


def pick(a,n):
    ans = set()
    for i in range(n):
        if isPick(a,n,a[i],i-1,i+1):
            ans.add(a[i])
    for i in range(n):
        if isDown(a,n,a[i],i-1,i+1):
            ans.add(a[i])
    return ans

def solve():
    T = int(sys.stdin.readline())
    for _ in range(T):
        n = int(sys.stdin.readline())
        a = list(map(int,sys.stdin.readline().split()))
        res = pick(a,n)
        ans = []
        for i in range(n):
            if a[i] in res:
                ans.append(a[i])
        print(len(ans))
        print(*ans)



if __name__ == "__main__":
    try:
        sys.stdin = open("input.txt","r")
    except :
        pass
    solve()











"""
5 1 9 3 8

 4 8 6 5





"""
