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].
#include <bits/stdc++.h>
using namespace std;
struct hash_pair {
  template <class T1, class T2>
  size_t operator()(const pair<T1, T2>& p) const {
    auto hash1 = hash<T1>{}(p.first);
    auto hash2 = hash<T2>{}(p.second);
    return hash1 ^ hash2;
  }
};
long long t;
long long mod = 1e9 + 7;
long long BigMod(long long base, long long pow, long long mod) {
  if (pow == 0 || base == 1)
    return 1;
  else if (pow == 1)
    return base;
  else if (pow % 2)
    return BigMod(base, pow - 1, mod) * base % mod;
  else {
    long long x = (BigMod(base, pow / 2, mod) % mod);
    x = (x * x) % mod;
    return x;
  }
}
vector<long long> v;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long a, b, c, d, e, f, g, i, j;
  long long n, m, k;
  cin >> t;
  while (t--) {
    cin >> n;
    vector<long long> ans;
    f = 0;
    g = -1;
    for (i = 0; i < n; i++) {
      cin >> a;
      v.push_back(a);
      if (i == 0) {
        ans.push_back(a);
        continue;
      }
      if (f == 0) {
        if (v[i] > v[i - 1]) {
          f = 1;
        } else if (v[i] < v[i - 1]) {
          f = -1;
        }
        continue;
      }
      if (f == 1 && v[i] < v[i - 1]) {
        ans.push_back(v[i - 1]);
        g = i;
        f = -1;
      } else if (f == -1 && v[i] > v[i - 1]) {
        ans.push_back(v[i - 1]);
        g = i;
        f = 1;
      }
    }
    if (ans.back() != v[n - 1] || ans.size() == 1) ans.push_back(v[n - 1]);
    cout << ans.size() << endl;
    for (i = 0; i < ans.size(); i++) {
      cout << ans[i] << ' ';
    }
    cout << endl;
    v.clear();
  }
  return 0;
}
