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:
# read test cases
testCases = int(input())

while testCases > 0:

    # get size of permutation
    size = int(input())

    # get permutation
    permutation = list(map(int, input().split()))

    # start subsequence with first element of permutation
    subsequence = [permutation[0]]

    # get if added element of subsequence is bigger then its following element
    if permutation[0] > permutation[1]:
        operation = 'bigger'
    else:
        operation = 'smaller'

    # iterate over permutation
    for i in range(1, size - 1):

        # operation remains the same -> skip to next element
        if (operation == 'bigger' and permutation[i] > permutation [i + 1]) or (operation == 'smaller' and permutation[i] < permutation [i + 1]):
            continue

        # add element to subsequence since it has biggest diff on ascending or
        # descinding sequence
        subsequence.append(permutation[i])

        # update next operation
        if operation == 'bigger':
            operation = 'smaller'
        else:
            operation = 'bigger'

    # add last element of sequence
    subsequence.append(permutation[-1])

    # print response
    print(len(subsequence))
    for i in range(len(subsequence)):
        if i == len(subsequence) - 1:
            print(str(subsequence[i]))
        else:
            print(str(subsequence[i]) + ' ', end = '')

    # decrement test cases
    testCases -= 1