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 Sat Jun 13 20:39:09 2020

@author: rishi
"""


t=int(input())
ans=[]
ans2=[]
for i in range(t):
    n=int(input())
    a=list(map(int,input().split()))
    inc=0
    dec=0
    num=0
    while True:
        if(a[num]!=a[num+1]):
            if(a[num]>a[num+1]):
                dec=1
                #ans.append(a[num])
                break
            else:
                #ans.append(a[num])
                inc=1
                break
        num+=1
    br=[a[0]]
    for j in range(n-1):
        
        if(a[j]==a[j+1]):
            continue
        else:
            if(inc and a[j]>a[j+1]):
                inc=0
                dec=1
                br.append(a[j])
                continue
            if(dec and a[j]<a[j+1]):
                dec=0
                inc=1
                br.append(a[j])
    br.append(a[n-1])
    ans.append(len(br))
    ans2.append(br)

for jj in range(t):
    print(ans[jj])
    for m in ans2[jj]:
        print(m,end=" ")