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>
void print(const T& v) {
  cerr << v << ' ';
}
template <typename T1, typename... T2>
void print(const T1& first, const T2&... rest) {
  print(first);
  print(rest...);
}
using vi = vector<int>;
using ll = long long;
using pii = pair<int, int>;
const int maxn = 1e5 + 5;
int n;
stack<int> s;
vector<int> ans;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  int a[maxn];
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 0; i <= (int)n - 1; i++) cin >> a[i];
    s.push(a[0]);
    int p = a[0];
    s.push(a[1]);
    for (int i = 2; i <= (int)n - 1; i++) {
      if (s.top() < a[i] && p < s.top())
        s.pop();
      else if (s.top() > a[i] && p > s.top())
        s.pop();
      p = s.top();
      s.push(a[i]);
    }
    cout << (int)s.size() << endl;
    ans.clear();
    while (!s.empty()) {
      ans.push_back(s.top());
      s.pop();
    }
    while (!ans.empty()) {
      cout << ans.back() << " ";
      ans.pop_back();
    }
    cout << endl;
  }
  return 0;
}