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].
"""
  ____          _      _____
 / ___|___   __| | ___|  ___|__  _ __ ___ ___  ___
| |   / _ \ / _` |/ _ \ |_ / _ \| '__/ __/ _ \/ __|
| |__| (_) | (_| |  __/  _| (_) | | | (_|  __/\__ \
 \____\___/ \__,_|\___|_|  \___/|_|  \___\___||___/

"""
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"""

import sys
import math
import collections
import operator as op
from collections import deque
from math import gcd

#sys.stdin = open('input.txt', 'r')
#sys.stdout = open('output.txt', 'w')

from functools import reduce
from sys import stdin, stdout, setrecursionlimit
setrecursionlimit(2**20)


def ncr(n, r):
    r = min(r, n - r)
    numer = reduce(op.mul, range(n, n - r, -1), 1)
    denom = reduce(op.mul, range(1, r + 1), 1)
    return numer // denom  # or / in Python 2


def isPowerOfTwo(x):
    return (x and (not(x & (x - 1))))


def factors(n):
    return (set(reduce(list.__add__, ([i, n // i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))

T = 1
T = int(stdin.readline())
for _ in range(T):
    # s = str(stdin.readline().strip('\n'))
    # a, b = list(map(int, stdin.readline().split()))
    # s = list(stdin.readline().strip('\n'))
    # b = list(stdin.readline().strip('\n'))
    n = int(stdin.readline())
    #n, x = list(map(int, stdin.readline().split()))
    p = list(map(int, stdin.readline().split()))
    ans = [p[0]]
    for i in range(1, n - 1):
        if p[i] > p[i + 1] and p[i] > p[i - 1]:
            ans.append(p[i])
        elif p[i] < p[i + 1] and p[i] < p[i - 1]:
            ans.append(p[i])
    ans.append(p[-1])
    print(len(ans))
    print(*ans)
