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;
int64_t factorial(int64_t n) {
  return (n == 1 || n == 0) ? 1 : n * factorial(n - 1);
}
int64_t power(int64_t x, int64_t y) {
  int64_t res = 1;
  while (y > 0) {
    if (y & 1) res = res * x;
    y = y >> 1;
    x = x * x;
  }
  return res;
}
int64_t powerMod(int64_t x, int64_t y, int64_t m) {
  if (y == 0) return 1;
  int64_t p = powerMod(x, y / 2, m) % m;
  p = (p * p) % m;
  return (y % 2 == 0) ? p : (x * p) % m;
}
int64_t modInverse(int64_t a, int64_t m) { return powerMod(a, m - 2, m); }
int64_t gcd(int64_t a, int64_t b) {
  if (a == 0) {
    return b;
  } else if (b == 0) {
    return a;
  } else {
    if (a > b) {
      return gcd(a % b, b);
    } else {
      return gcd(b % a, a);
    }
  }
}
void pre() {}
void solve() {
  int64_t n = 0;
  cin >> n;
  vector<int64_t> v(n);
  for (int64_t i = 0; i < n; i++) cin >> v[i];
  vector<int64_t> ans;
  for (int64_t i = 0; i < n; i++) {
    if (i == 0 || i == n - 1)
      ans.push_back(v[i]);
    else if (v[i] > v[i - 1] && v[i] > v[i + 1])
      ans.push_back(v[i]);
    else if (v[i] < v[i - 1] && v[i] < v[i + 1])
      ans.push_back(v[i]);
  }
  cout << ans.size();
  cout << "\n";
  for (auto itr : ans) cout << itr << " ";
  ;
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int64_t t = 1;
  cin >> t;
  pre();
  while (t--) {
    solve();
  }
  return 0;
}