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;
int main() {
  int t, n, i, j, k;
  int* a;
  vector<int> in;
  cin >> t;
  while (t--) {
    cin >> n;
    a = new int[n];
    for (i = 0; i < n; i++) {
      cin >> a[i];
    }
    i = 0;
    while (i < n - 1) {
      if (a[i] < a[i + 1] && i < n - 1) {
        k = i;
        j = i + 1;
        while (j < n && a[k] < a[j]) {
          k++;
          j++;
        }
        if (in.empty() || in[in.size() - 1] != i) {
          in.push_back(i);
        }
        in.push_back(k);
        i = k;
      }
      if (a[i] > a[i + 1] && i < n - 1) {
        k = i;
        j = i + 1;
        while (j < n && a[k] > a[j]) {
          k++;
          j++;
        }
        if (in.empty() || in[in.size() - 1] != i) {
          in.push_back(i);
        }
        in.push_back(k);
        i = k;
      }
    }
    cout << in.size() << endl;
    for (i = 0; i < in.size(); i++) {
      cout << a[in[i]] << " ";
    }
    cout << "\n";
    in.clear();
  }
}