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 ca1(n,lis):
    l1=0
    l2=0
    for i in range(1, n):
        if l1 == 0:
            if lis[i] >= lis[i - 1]:
                pass
            else:
                nu1.append(lis[i - 1])
                l1 = 1
                l2 = 0
        else:
            if lis[i] <= lis[i - 1]:
                pass
            else:
                nu1.append(lis[i - 1])
                l1 = 0
                l2 = 1
    nu1.append(lis[-1])
    return nu1
def ca2(n,lis):
    l1 = 0
    l2 = 0
    for i in range(1, n):
        if l1 == 0:
            if lis[i] <= lis[i - 1]:
                pass
            else:
                nu2.append(lis[i - 1])
                l1 = 1
                l2 = 0
        else:
            if lis[i] >= lis[i - 1]:
                pass
            else:
                nu2.append(lis[i - 1])
                l1 = 0
                l2 = 1
    nu2.append(lis[-1])
    return nu2
t=int(input())
for tc in range(t):
    n=int(input())
    lis=list(map(int,input().split()))
    nu1=[lis[0]]
    nu2=[lis[0]]
    re1=nu1
    re2=nu2
    if lis[0]<lis[1]:
        re1=ca1(n,lis)
    else:
        re2=ca2(n,lis)
    if len(re1)>len(re2):
        l1=len(re1)
        print(l1)
        for i in re1:
            print(i,end=" ")
        print()
    else:
        l2 = len(re2)
        print(l2)
        for i in re2:
            print(i, end=" ")
        print()