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].
"Codeforces Round #649 (Div. 2)"
"A. XXXXX"
# y=int(input())
# ans=[]
# for i in range(y):
#     al=input().split()
#     m = list(map(int, al))
#     n=m[0]
#     k=m[1]
#     al=input().split()
#     m = list(map(int, al))
#     s=sum(m)
#     s1=s
#     l=n
#     z=0
#     x=n-1
#     while s%k==0:
#         s=s-m[z]
#         z+=1
#         if z>n-2:
#             z=-1
#             break
#     while s1%k==0:
#         s1=s1-m[x]
#         x-=1
#         if x<1:
#             x=-1
#             break    
#     if x==-1 and z==-1:
#         ans.append(-1)
#     elif x==-1:
#         ans.append(n-z)
#     elif z==-1:
#         ans.append(x+1)
#     else:
#         ans.append(max(n-z,x+1))                         

# for i in ans:
#     print(i)
"B. Most socially-distanced subsequence"
# def score(a):
#     sc=0
#     for i in range(1,len(a)-1):
#         sc+=abs(a[i]-a[i+1])+abs(a[i]-a[i-1])
#     if len(a)>2:
#         sc+=abs(a[1]-a[2])+abs(a[-2]-a[-1])
#     else:
#         sc+=abs(a[1]-a[2])
#     return sc            

y=int(input())
ans=[]
for i in range(y):
    n=int(input())
    al=input().split()
    m = list(map(int, al))
    an=[]
    an.append(m[0])
    for i in range(1,n-1):
        if (abs(m[i]-m[i+1])+abs(m[i]-m[i-1]))>(abs(m[i+1]-m[i-1])):
            an.append(m[i])
    an.append(m[-1])
    ans.append(an)

for i in ans:
    print(len(i))
    for j in i:
        print(j,end=" ")
    print()                
