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:
for _ in range(int(input())):
    n=int(input())
    l = list(map(int, input().split()))
    temp=[]
    count=0
    for i in range(n-1):
        temp.append(l[i+1]-l[i])
    ans=[]
    flag=0
    if temp[0]>0:
        flag=1
    else:
        flag=0
    ans.append(l[0])
    count+=1
    # print(temp)
    flag2=0
    for i in range(1,len(temp)):
        if temp[i]>=0 and flag==1:
            continue
        elif temp[i]<0 and flag==0:
            continue
        else:
            if temp[i]<0:
                flag=0
            else:
                flag=1
            count+=1
            ans.append(l[i])
            if i==n-1:
                flag2=1
    if flag2==0:
        count+=1
        ans.append(l[n-1])
    print(count)
    for i in range(len(ans)):
        print(ans[i],end=" ")
    print()