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;
void solver() {
  int n;
  cin >> n;
  vector<int> p(n);
  for (int &x : p) cin >> x;
  vector<int> ans;
  ans.push_back(p[0]);
  for (int i = 0; i < n; ++i) {
    while (i + 1 < n && p[i] > p[i + 1]) ++i;
    if (p[i] != ans.back()) ans.push_back(p[i]);
    while (i + 1 < n && p[i] < p[i + 1]) ++i;
    if (p[i] != ans.back()) ans.push_back(p[i]);
  }
  n = ans.size();
  cout << n << '\n';
  for (int i = 0; i < n; ++i) cout << ans[i] << " \n"[i == n - 1];
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) solver();
  return 0;
}