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:
def triplSum(a1, a2, a3):
    return abs(a1-a2) + abs(a2-a3)
def doublelSum(a1, a2):
    return abs(a1-a2)
for _ in range(int(input())):
    n = int(input())
    a = list(map(int, input().split()))
    s = 0
    i = 0
    i1 = 1
    b = [a[0]]
    while True:
        # print(i, i1)
        if i1 + 1 == n-1:
            if triplSum(a[i], a[i1], a[i1 + 1]) > doublelSum(a[i], a[i1 + 1]):
                s += triplSum(a[i], a[i1], a[i1 + 1])
                b.append(a[i1])
                b.append(a[i1 + 1])
            else:
                s += doublelSum(a[i], a[i1+1])
                b.append(a[i1 + 1])
            break
        elif i1 == n-1:
            if b[-1] == a[i]:
                b.append(a[i1])
            break
        else:
            if triplSum(a[i], a[i1], a[i1+1]) > doublelSum(a[i], a[i1+1]):
                # s += triplSum(a[i], a[i1], a[i1+1])
                b.append(a[i1])
                i = i1
                i1 = i + 1
            else:
                i1 += 1
    print(len(b))
    print(*b)