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;
long long int power(long long int a, long long int b) {
  if (b == 0) return 1;
  if (b % 2 == 0) return (power((a * a) % 998244353, b / 2)) % 998244353;
  return (a * (power((a * a) % 998244353, b / 2)) % 998244353) % 998244353;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int modInv(int n) { return power(n, 1000000007 - 2); }
int nCr(int n, int r, int p = 1000000007) {
  if (r == 0) return 1;
  int Fact[n + 1];
  Fact[0] = 1;
  for (int i = 1; i <= n; i++) {
    Fact[i] = (Fact[i - 1] * i) % p;
  }
  return (((Fact[n] * modInv(Fact[r])) % 1000000007) *
          (modInv(Fact[n - r]) % 1000000007)) %
         1000000007;
}
int main() {
  int t = 1;
  cin >> t;
  while (t--) {
    int n, i, j;
    cin >> n;
    int a[n];
    for (i = 0; i < n; i++) {
      cin >> a[i];
    }
    vector<int> ans;
    ans.push_back(a[0]);
    ans.push_back(a[1]);
    int q = a[1] - a[0];
    for (i = 2; i < n; i++) {
      int r = a[i] - a[i - 1];
      if ((r > 0 && q > 0) || (r < 0 && q < 0))
        ans.back() = a[i];
      else
        ans.push_back(a[i]);
      q = r;
    }
    cout << ans.size() << "\n";
    for (i = 0; i < ans.size(); i++) {
      cout << ans[i] << " ";
    }
    cout << "\n";
  }
}
