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:
import sys
import bisect as b
import math
from collections import defaultdict as dd
input=sys.stdin.readline
mo=10**9+7
def cin():
    return map(int,sin().split())
def ain():            
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input())
def pref(a,n):             #preSum array
    pre=[0]*n
    pre[0]=a[0]
    for i in range(1,n):
        pre[i]=a[i]+pre[i-1]
    return pre
for _ in range(inin()):
    n=inin()
    l=ain()
    s=1
    f=f1=0
    st=l[0]
##    st2=l[0]
    ans=[]
##    a=[l[0]]
##    end1=end2=l[-1]
    c=2
    for i in range(0,n-1):
        if(l[i+1]>l[i] ):
            if(f==0):
                ans+=[st]
                s+=1
            f=1
            f1=0
            st=l[i+1]
##            print(st,i)
        elif(l[i+1]<l[i]):
            if(f1==0):
##                print('ans')
                ans+=[st]
                s+=1
            st=l[i+1]
            f1=1
            f=0
##        print(st,end1,end2,ans,i)
##    if(f1==0):
##        ans+=[st,end2]
##    else:
##        ans+=[st,end1]
    print(s)
    x=ans+[st]
    print(*x)