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>
T modpow(T base, T exp, T modulus) {
  base %= modulus;
  T result = 1;
  while (exp > 0) {
    if (exp & 1) result = (result * base) % modulus;
    base = (base * base) % modulus;
    exp >>= 1;
  }
  return result;
}
template <typename T>
T modsub(T a, T b, T mod) {
  return ((a - b) % mod + mod) % mod;
}
template <typename T>
T moddiv(T a, T b, T mod) {
  return a * modpow(b, mod - 2, mod) % mod;
}
template <typename T>
inline T gcd(T a, T b) {
  while (b != 0) swap(b, a %= b);
  return a;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(10);
  cout << fixed;
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    vector<int> num(n, 0);
    for (int i = 0; i < n; i += 1) {
      cin >> num[i];
    }
    vector<int> ans;
    for (int i = 0; i < n; i++) {
      if (i == 0 || i == n - 1 ||
          (num[i - 1] < num[i]) != (num[i] < num[i + 1])) {
        ans.emplace_back(num[i]);
      }
    }
    cout << ans.size() << "\n";
    for (auto i : ans) cout << i << " ";
    cout << "\n";
  }
  return 0;
}