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()))
    k = [] # new list after removing
    c = ''

    if l[0] < l[1]:
        c = 'i' # increasing
    else:
        c = 'd' # decreasing

    # x = 1 # start of sequence
    # count = 0
    # for i in range(0,n-1):
    #     if (l[i] <= l[i+1] and c == 'd') or (l[i] >= l[i+1] and c == 'i'): # change in increse/decrease
    #         # sequence: 1 -> i
    #         k.append(str(x)) # 1st element
    #         k.append(str(i)) # last element
    #         x = i # reset first element to last element of previous sequence
    #         count += 1
    #
    # if count == 0:
    #     print(2)
    #     print(l[0],l[-1])
    k.append(l[0])
    k.append(l[1])
    for i in range(1,n-1):
        if l[i] <= l[i+1]:
            if c == 'i':
                k[len(k)-1] = l[i+1]
            else:
                k.append(l[i+1])
            c = 'i'
        elif l[i] >= l[i+1]:
            if c == 'd':
                k[len(k)-1] = l[i+1]
            else:
                k.append(l[i+1])
            c = 'd'

    print(len(k))
    print(' '.join(map(str,k)))

# 3 7 2 4 5 6 1
#  4 5 2 1 1 5  = 18
# 3 7 2 x x 6 1