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 T>
inline ostream &operator<<(ostream &_out, const vector<T> &_v) {
  if (_v.empty()) return _out;
  _out << _v.front();
  for (auto _it = ++_v.begin(); _it != _v.end(); ++_it) _out << ' ' << *_it;
  return _out;
}
template <typename T>
inline istream &operator>>(istream &_in, vector<T> &_v) {
  for (auto &_i : _v) _in >> _i;
  return _in;
}
template <typename T>
inline ostream &operator<<(ostream &_out, const set<T> &_s) {
  if (_s.empty()) return _out;
  _out << *_s.begin();
  for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) _out << ' ' << *_it;
  return _out;
}
template <typename T, typename U>
inline ostream &operator<<(ostream &_out, const map<T, U> &_m) {
  if (_m.empty()) return _out;
  _out << '(' << _m.begin()->first << ": " << _m.begin()->second << ')';
  for (auto _it = ++_m.begin(); _it != _m.end(); ++_it)
    _out << ", (" << _it->first << ": " << _it->second << ')';
  return _out;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
inline void solve() {
  int n;
  cin >> n;
  vector<int> p(n);
  cin >> p;
  int begin_of_current_segment = 0;
  bool increase = p[1] > p[0];
  vector<bool> mask(n, false);
  for (int i = 2; i < n; ++i) {
    if (p[i] > p[i - 1]) {
      if (!increase) {
        increase = true;
        mask[begin_of_current_segment] = true;
        mask[i - 1] = true;
        begin_of_current_segment = i - 1;
      }
    } else {
      if (increase) {
        increase = false;
        mask[begin_of_current_segment] = true;
        mask[i - 1] = true;
        begin_of_current_segment = i - 1;
      }
    }
  }
  mask[begin_of_current_segment] = true;
  mask.back() = true;
  cout << accumulate(mask.begin(), mask.end(), 0) << "\n";
  for (int i = 0; i < n; ++i) {
    if (mask[i]) cout << p[i] << " ";
  }
  cout << "\n";
}
inline void tsolve() {
  int t;
  cin >> t;
  while (t--) solve();
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  tsolve();
  return 0;
}