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:
# -*- coding: utf-8 -*-
"""
Created on Mon Jun 15 18:22:19 2020

@author: sun
"""
def tostr(A):
    return [*map(str, A)]
def toint(A):
    return [*map(int, A)]
def solve(A):
    if len(A) <= 2:
        return A
    i,j,k  = 0,1,2
    ret = []
    while k<len(A):
        if (A[i]-A[j]) *(A[k]-A[j]) > 0:
            ret.append(A[i])
            i,j,k = j,k,k+1
        else:
            i,j,k = i,k,k+1
    ret.append(A[i])
    ret.append(A[j])
    return ret

from os import sys
ncases = int(sys.stdin.readline().strip())
for _ in range(ncases):
    n = sys.stdin.readline()
    A = sys.stdin.readline().split()
    ret = solve(toint(A))
    print(len(ret))
    print(" ".join(tostr(ret)))