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;
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
const double pi = 3.14159265359;
const long long mod = 1000000007;
const long long modc = 998244353;
long long inline power(long long a, long long b, long long p) {
  a %= p;
  long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = (ans * a) % p;
    a = (a * a) % p;
    b >>= 1;
  }
  return ans;
}
long long mod_inv(long long n, long long p) { return power(n, p - 2, p); }
bool inline isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
template <typename... T>
void W(T &&...args) {
  ((cout << fixed << setprecision(9) << args << " "), ...);
  cout << endl;
}
template <typename... T>
void R(T &&...args) {
  ((cin >> args), ...);
}
long long ston(string s) {
  long long a = 0;
  reverse(s.begin(), s.end());
  long long pow = 1;
  for (int i = 0; i < s.size(); i++) {
    a += pow * (s[i] - '0');
    pow *= 10;
  }
  return a;
}
template <class T>
string to_binary(T n) {
  string binaryNum;
  int i = 0;
  while (n > 0) {
    binaryNum.push_back((n % 2) + '0');
    n = n / 2;
    i++;
  }
  reverse(binaryNum.begin(), binaryNum.end());
  return binaryNum;
}
template <class T>
bool perfectSq(T n) {
  if (sqrt(n) == floor(sqrt(n)))
    return true;
  else
    false;
}
template <class T>
void wv(T &v) {
  for (auto x : v) cout << x << " ";
  cout << endl;
}
template <class T>
void rv(T &v) {
  for (int i = 0; i < (int)v.size(); i++) cin >> v[i];
}
void solve() {
  int i, j;
  long long sum = 0, prod = 1;
  long long maxl = LONG_MIN, minl = LONG_MAX;
  int maxi = INT_MIN, mini = INT_MAX;
  int n;
  R(n);
  vector<int> v(n);
  rv(v);
  vector<int> ans;
  ans.push_back(v[0]);
  for (i = 1; i < n - 1; i++) {
    if ((v[i] > v[i - 1] && v[i] > v[i + 1]) ||
        (v[i] < v[i - 1] && v[i] < v[i + 1]))
      ans.push_back(v[i]);
  }
  ans.push_back(v[n - 1]);
  cout << (int)ans.size() << endl;
  wv(ans);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}
