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 Wed Jun 10 22:06:01 2020

@author: Mridul Garg
"""
from collections import deque

q = int(input())
for _ in range(q):
    n = int(input())
    A = list(map(int, input().split(" ")))
    
    temp = A[0]
    if A.count(temp) == n:
        print(2)
        print(temp, temp)
        continue
    
    l, r = 0, 1
    for i in range(1, n):
        if A[i] == A[i-1]:
            continue
        elif A[i] > A[i-1]:
            l = i-1
            r = i
            inc = True
            break
        else:
            l = i-1
            r = i
            inc = False
            break
    
    start = r+1
    ans = []
    ans.append(A[l])
    for i in range(start, n-1):
        if inc:
            if A[i] >= A[i-1]:
                r = i
            else:
#                ans.append(A[l])
                ans.append(A[r])
                l = i-1
                r = i
                inc = False
        else:
            if A[i] <= A[i-1]:
                r = i
            else:
#                ans.append(A[l])
                ans.append(A[r])
                l = i-1
                r = i
                inc = True
    
    
    i = n-1
#    ans.append(A[l])
    if inc:
        if A[i] > A[i-1]:
            ans.append(A[i])
        if A[i] < A[i-1]:
            ans.append(A[r])
            ans.append(A[i])
    else:
        if A[i] < A[i-1]:
            ans.append(A[i])
        if A[i] > A[i-1]:
            ans.append(A[r])
            ans.append(A[i])
    
    print(len(ans))
    for i in ans:
        print(i, end = " ")
    print()