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 read() {
  long long sign = 1, x = 0;
  char s = getchar();
  while (s > '9' || s < '0') {
    if (s == '-') sign = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  return x * sign;
}
void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x / 10) write(x / 10);
  putchar(x % 10 + '0');
}
int n, k;
int a[(500000 + 5)];
bool judge(int i) { return a[i] == a[i - 1] + 1; }
bool judge2(int i) { return a[i] == a[i - 1] - 1; }
int main() {
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 1; i <= n; ++i) {
      cin >> a[i];
    }
    vector<int> ans, ans1;
    ans.clear();
    ans.push_back(a[1]);
    for (int i = 2; i <= n; ++i) {
      if (a[i] > a[i - 1]) {
        while (a[i] < a[i + 1] && i < n) {
          ++i;
        }
        ans.push_back(a[i]);
      } else {
        while (a[i] > a[i + 1] && i < n) {
          ++i;
        }
        ans.push_back(a[i]);
      }
    }
    cout << ans.size() << endl;
    for (auto it : ans) {
      cout << it << " ";
    }
    cout << endl;
  }
  return 0;
}