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:
/**
 * @author Finn Lidbetter
 */
import java.util.*;
import java.io.*;
import java.awt.geom.*;

public class TaskB {
  public static void main(String[] args) throws IOException {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringBuilder sb = new StringBuilder();
    
    int nTests = Integer.parseInt(br.readLine());
    for (int test=0; test<nTests; test++) {
      int n = Integer.parseInt(br.readLine());
      int[] arr = new int[n];
      String[] s = br.readLine().split(" ");
      for (int i=0; i<n; i++) {
        arr[i] = Integer.parseInt(s[i]);
      }
      Stack<Integer> stk = new Stack<Integer>();
      stk.push(arr[0]);
      int dir = 0;
      for (int i=1; i<n; i++) {
        if (dir==0) {
          if (arr[i]>stk.peek()) {
            dir = 1;
          } else {
            dir = -1;
          }
          stk.push(arr[i]);
        } else if (dir==1) {
          if (arr[i]>stk.peek()) {
            stk.pop();
            stk.push(arr[i]);
          } else {
            dir = -1;
            stk.push(arr[i]);
          }
        } else {
          if (arr[i]<stk.peek()) {
            stk.pop();
            stk.push(arr[i]);
          } else {
            dir = 1;
            stk.push(arr[i]);
          }
        }
      }
      int len = stk.size();
      int[] output = new int[len];
      for (int i=0; i<len; i++) {
        output[len-i-1] = stk.pop();
      }
      sb.append(len+"\n");
      sb.append(output[0]);
      for (int i=1; i<len; i++) {
        sb.append(" "+output[i]);
      }
      sb.append("\n");
    }
    System.out.print(sb);
  }
  

}

class MyUtils {
  static void shuffleArray(int[] arr) {
    int n = arr.length;
    Random rnd = new Random();
    for(int i=0; i<n; i++) {
      int tmp = arr[i];
      int randomPos = i + rnd.nextInt(n-i);
      arr[i] = arr[randomPos];
      arr[randomPos] = tmp;
    }
  }
  static void shuffleArray(long[] arr) {
    int n = arr.length;
    Random rnd = new Random();
    for(int i=0; i<n; i++) {
      long tmp = arr[i];
      int randomPos = i + rnd.nextInt(n-i);
      arr[i] = arr[randomPos];
      arr[randomPos] = tmp;
    }
  }
}