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;
int _gcd(int a, int b) { return (b == 0 ? a : _gcd(b, a % b)); }
bool cmp(int a, int b) { return a > b; }
int dx[3] = {1, 0, 1};
int dy[3] = {0, 1, 1};
bool check(string a, string b, string c) {
  if (a == b && b == c) return true;
  if (a > b) swap(a, b);
  if (b > c) swap(b, c);
  if (a > b) swap(a, b);
  if (a[1] == b[1] && b[1] == c[1]) {
    if (a[0] + 1 == b[0] && b[0] + 1 == c[0]) return true;
  }
  return false;
}
long long int ncr(long long int n, long long int k) {
  long long int res = 1;
  for (long long int i = n; i > n - k; --i) {
    res *= i;
  }
  for (long long int i = 2; i <= k; ++i) {
    res /= i;
  }
  return res;
}
const int mod = 1e9 + 7;
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> v(n + 1, 0);
  for (long long int i = 1; i <= n; ++i) {
    cin >> v[i];
  }
  vector<long long int> ans;
  for (long long int i = 1; i <= n; ++i) {
    if (i == 1 || i == n || (v[i - 1] < v[i]) != (v[i] < v[i + 1]))
      ans.push_back(v[i]);
  }
  cout << ans.size() << endl;
  for (auto x : ans) {
    cout << x << " ";
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
