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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long ans[100005], a[100005], p;
int main() {
  long long t;
  scanf("%lld", &t);
  while (t--) {
    long long n;
    scanf("%lld", &n);
    for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
    long long f = 0, p = 0;
    ans[++p] = a[1];
    for (int i = 2; i <= n; i++) {
      if (a[i] >= a[i - 1]) {
        if (!f)
          f = 1;
        else if (f == 1)
          ;
        else {
          ans[++p] = a[i - 1], f = 1;
        }
      } else {
        if (!f)
          f = 2;
        else if (f == 2)
          ;
        else {
          ans[++p] = a[i - 1], f = 2;
        }
      }
    }
    if (f) ans[++p] = a[n];
    printf("%lld\n", p);
    for (int i = 1; i <= p; i++) printf("%lld ", ans[i]);
    printf("\n");
  }
}