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;
template <typename S, typename T>
ostream& operator<<(ostream& out, pair<S, T> const& p) {
  out << '(' << p.first << ", " << p.second << ')';
  return out;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> const& v) {
  long long l = v.size();
  for (long long i = 0; i < l - 1; i++) out << v[i] << ' ';
  if (l > 0) out << v[l - 1];
  return out;
}
template <typename T>
void trace(const char* name, T&& arg1) {
  cout << name << " : " << arg1 << "\n";
}
template <typename T, typename... Args>
void trace(const char* names, T&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  trace(comma + 1, args...);
}
void solve() {
  int n;
  cin >> n;
  vector<int> arr(n);
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  vector<int> ans;
  ans.push_back(arr[0]);
  ans.push_back(arr[1]);
  bool inc = arr[1] > arr[0];
  for (int i = 2; i < n; i++) {
    if (arr[i] > arr[i - 1]) {
      if (!inc) {
        ans.push_back(arr[i]);
        inc = true;
      } else {
        ans.back() = arr[i];
      }
    } else {
      if (inc) {
        ans.push_back(arr[i]);
        inc = false;
      } else {
        ans.back() = arr[i];
      }
    }
  }
  cout << ans.size() << "\n";
  for (auto i : ans) {
    cout << i << " ";
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}