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;
long long int mod = 1000000007;
long long int rmod = 1000000007;
long long int power(long long int a, long long int b) {
  long long int value = 1;
  while (b) {
    if (b & 1) {
      value *= a;
    }
    a *= a;
    b /= 2;
  }
  return value;
}
bool isPrime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    long long int a[n];
    for (long long int i = 0; i < n; i++) {
      cin >> a[i];
    }
    long long int x = 0;
    vector<long long int> v;
    v.push_back(a[0]);
    if (a[0] > a[1]) {
      x = 1;
    } else if (a[0] < a[1]) {
      x = -1;
    } else {
      x = 0;
    }
    long long int x1 = 0;
    for (int i = 2; i < n; i++) {
      if (a[i - 1] > a[i]) {
        x1 = 1;
      } else if (a[i - 1] < a[i]) {
        x1 = -1;
      } else {
        x1 = 0;
      }
      if (x != x1 && x != 0) {
        v.push_back(a[i - 1]);
      }
      swap(x, x1);
    }
    if (a[n - 1] != a[n - 2]) {
      v.push_back(a[n - 1]);
    }
    if (v.size() >= 2) {
      cout << v.size() << endl;
      for (auto i : v) {
        cout << i << " ";
      }
      cout << endl;
    } else {
      cout << v.size() + 1 << endl;
      cout << v[0] << " " << a[n - 1];
      cout << endl;
    }
  }
  return 0;
}