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;
const long long int mod = 1e9 + 7;
const long long int N = 1e6 + 100;
long long int pr[N] = {0};
long long int SEr() {
  for (long long int i = 2; i < N; i++) {
    if (pr[i] == 0) {
      for (long long int j = 2 * i; j < N; j += i) pr[j]++;
    }
  }
  return 0;
}
bool isPrime(long long int n) {
  if (pr[n] == 0)
    return true;
  else
    return false;
}
long long int power(long long int a, long long int b) {
  a %= mod;
  long long int res = 1;
  while (b > 0) {
    if (b % 2) {
      res *= a;
      res %= mod;
    }
    a *= a;
    a %= mod;
    b /= 2;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) {
    return a;
  } else {
    return gcd(b, a % b);
  }
}
long long int lcm(long long int a, long long int b) {
  return ((a / gcd(a, b)) * b);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int T = 1;
  cin >> T;
  while (T--) {
    long long int n, i;
    cin >> n;
    vector<int> p(n);
    for (i = 0; i < n; i++) cin >> p[i];
    vector<int> ans;
    for (i = 0; i < n; i++) {
      if (i == 0 || i == n - 1 || (p[i] < p[i - 1] == p[i] < p[i + 1]))
        ans.push_back(p[i]);
    }
    cout << ans.size();
    cout << "\n";
    for (auto a : ans) {
      cout << a << " ";
    }
    cout << "\n";
  }
  return 0;
}