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:
import re
import sys
from bisect import bisect, bisect_left, insort, insort_left
from collections import Counter, defaultdict, deque
from copy import deepcopy
from decimal import Decimal
from itertools import (
    accumulate, combinations, combinations_with_replacement, groupby,
    permutations, product)
from math import (acos, asin, atan, ceil, cos, degrees, factorial, gcd, hypot,
                  log2, pi, radians, sin, sqrt, tan)
from operator import itemgetter, mul
from string import ascii_lowercase, ascii_uppercase, digits


def inp():
    return(int(input()))


def inlist():
    return(list(map(int, input().split())))


def instr():
    s = input()
    return(list(s[:len(s)]))


def invr():
    return(map(int, input().split()))


t = inp()

for _ in range(t):
    n = inp()
    a = inlist()

    res = []
    res.append(a[0])
    flip = 0
    if a[1] < a[0]:
        flip = 1
    for i in range(1, n):
        if flip == 0 and a[i] < a[i-1]:
            res.append(a[i-1])
            flip = 1
        elif flip == 1 and a[i] >= a[i-1]:
            res.append(a[i-1])
            flip = 0
    res.append(a[n-1])
    print(len(res))
    print(*res)