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>
const int mod = 998244353;
using namespace std;
template <typename T>
class fenwick_tree {
  vector<T> t;
  vector<T> arr;
  string convert_to_binary(T num) {
    string s = "";
    for (; num; num >>= 1) {
      s.push_back((num & 1) + '0');
    }
    s.push_back('0');
    reverse(s.begin(), s.end());
    return s;
  }
  void build() {
    for (size_t i = 0; i < arr.size(); ++i) {
      inc(i, arr[i]);
    }
  }

 public:
  fenwick_tree(vector<T>& arr) {
    this->arr = arr;
    t.resize(arr.size());
    build();
  }
  fenwick_tree(vector<T>&& arr) {
    this->arr = move(arr);
    t.resize(this->arr.size());
    build();
  }
  void inc(int i, T val) {
    for (; i < t.size(); i = i | (i + 1)) {
      t[i] += val;
    }
  }
  T get(int i) {
    T num = 0;
    for (; i >= 0; i = (i & (i + 1)) - 1) {
      num += t[i];
    }
    return num;
  }
  T sum(int i, int j) { return get(j) - get(i - 1); }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> arr(n);
    for (auto& i : arr) {
      cin >> i;
    }
    vector<int> ans;
    ans.push_back(arr[0]);
    for (int i = 1; i < arr.size() - 1; ++i) {
      if ((arr[i] > arr[i - 1] && arr[i] > arr[i + 1]) ||
          (arr[i] < arr[i - 1] && arr[i] < arr[i + 1])) {
        ans.push_back(arr[i]);
      }
    }
    ans.push_back(arr.back());
    cout << ans.size() << endl;
    for (const auto& i : ans) {
      cout << i << " ";
    }
    cout << endl;
  }
  return 0;
}