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;
const long long mod = 1000000007;
long double pi = 3.14159265358979323846;
clock_t time_p = clock();
void rtime() {
  time_p = clock() - time_p;
  cerr << "Time Taken : " << fixed << (float)(time_p) / CLOCKS_PER_SEC << "\n";
}
bool isPrime(long long n) {
  for (long long i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
long long factorial(long long n) {
  return (n == 1 || n == 0) ? 1 : n * factorial(n - 1);
}
long long power(long long x, long long y) {
  long long res = 1;
  x = x;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res % mod;
}
long long ncr(long long n, long long r) {
  long long res = 1;
  if (r > n) return 0;
  if (r > n - r) r = n - r;
  for (long long i = 0; i < r; i++) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) { return (a / gcd(a, b) * b); }
long long max(long long a, long long b) {
  long long ans = a > b ? a : b;
  return ans;
}
long long min(long long a, long long b) {
  long long ans = a < b ? a : b;
  return ans;
}
long long n, m;
long long corner(long long i, long long j) {
  if ((i == 0 && (j == 0 || j == m - 1)) ||
      (i == n - 1 && (j == 0 || j == m - 1)))
    return 1;
  return 0;
}
long long three(long long i, long long j) {
  if ((i == 0 || i == n - 1) && (j > 0 && j < m - 1))
    return 1;
  else if ((j == 0 || j == m - 1) && (i > 0 && i < n - 1))
    return 1;
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t, n;
  cin >> t;
  while (t--) {
    cin >> n;
    long long a[n];
    for (long long i = 0; i < n; i++) cin >> a[i];
    vector<long long> ans;
    for (long long i = 0; i < n; i++) {
      if (i == 0 || i == n - 1)
        ans.push_back(a[i]);
      else if ((a[i] > a[i - 1] && a[i] > a[i + 1]) ||
               (a[i] < a[i - 1] && a[i] < a[i + 1]))
        ans.push_back(a[i]);
    }
    long long k = ans.size();
    cout << k << "\n";
    for (long long i = 0; i < k; i++) cout << ans[i] << " ";
    cout << "\n";
  }
  rtime();
}
