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.*;
import java.io.*;
public class Main1{
  public static void main(String[] args)throws IOException{
    Scanner sc = new Scanner(System.in);
    int a,b,c,d,e,f,g,h;
    a = sc.nextInt();
    for(b = 0;b < a;b++){
      c = sc.nextInt();
      int x[] = new int[c];
      for(d = 0;d < c;d++){
        x[d] = sc.nextInt();
      }
      //Arrays.sort(x);
      ArrayList<Integer> y  = new ArrayList<Integer>();
      //int z[] = new int [c];
      
      y.add(x[0]);
      y.add(x[1]);
      for(d = 2;d < c;d++){
        e = Math.abs(x[d] - y.get(y.size()- 1));
        f = Math.abs(y.get(y.size()- 1) - y.get(y.size()- 2));
        g = Math.abs(x[d] - y.get(y.size()- 2));
        //System.out.println(x[d]+" "+y.get(y.size()- 2)+" "+y.get(y.size()- 1));
        //System.out.println(g+" "+e+" "+f);
        if(g >= (e+f)){
        y.remove(y.size()- 1);
        }
        y.add(x[d]);
      }
      System.out.println(y.size());
      for(d = 0;d < y.size()-1;d++){
        System.out.print(y.get(d)+" ");
      }
      System.out.println(y.get(y.size()-1));
    }
  }
}