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 gcd(long long a, long long b) {
  while (a && b) {
    if (a > b)
      a = a % b;
    else
      b = b % a;
  }
  return a + b;
}
long long fact(int n) {
  if (n == 0) return 1;
  return fact(n - 1) * n;
}
long long lcm(long long a, long long b) {
  long long temp = gcd(a, b);
  return temp ? (a / temp * b) : 0;
}
bool isPolindrom(string s) {
  for (int i = 0; i < s.length() / 2; ++i) {
    if (s[i] != s[s.length() - 1 - i]) return false;
  }
  return true;
}
void sch(map<int, map<int, int>> &a, int u, vector<int> &used) {
  for (auto it = a[u].begin(); it != a[u].end(); ++it) {
    if (!used[it->first]) {
      used[it->first] = 1;
      sch(a, it->first, used);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    int n, k = 0;
    cin >> n;
    int a[n];
    for (int i = 0; i < n; ++i) cin >> a[i];
    for (int i = 1, j = 0; i < n - 1; ++i) {
      if (abs(a[j] - a[i + 1]) == abs(a[i] - a[j]) + abs(a[i] - a[i + 1])) {
        a[i] = 0;
        ++k;
      } else
        j = i;
    }
    cout << n - k << endl;
    for (int i = 0; i < n; ++i) {
      if (a[i]) cout << a[i] << " ";
    }
    cout << endl;
  }
  return 0;
}