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 JAVA):

Solution:
import java.util.*;
 
public class Solution{
    public static void main(String []args){
    Scanner sc=new Scanner(System.in);
    int t=sc.nextInt();
    for(int l=0;l<t;l++){
      int n=sc.nextInt();
      long a[]=new long[n];
      long b[]=new long[n];
      int i;
      for(i=0;i<n;i++)
      a[i]=sc.nextLong();
      int cnt=0;
      b[cnt]=a[0];
      cnt++;
      for(i=1;i<n-1;i++){
        if(a[i]-a[i-1]>0 
          && a[i]-a[i+1]>0){
          b[cnt]=a[i];
          cnt++;
          }
        else if(a[i]-a[i-1]<0 
          && a[i]-a[i+1]<0){
          b[cnt]=a[i];
          cnt++;
          }
      }
      b[cnt]=a[n-1];
      cnt++;
      System.out.println(cnt);
      for(i=0;i<cnt;i++)
      System.out.print(b[i]+" ");
      System.out.println();
    }
  }
}