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;
const int N = 1e5 + 10, mod = 1e9 + 7;
const double eps = 1e-10;
template <typename T>
void read(T &x) {
  x = 0;
  char ch = getchar();
  long long first = 1;
  while (!isdigit(ch)) {
    if (ch == '-') first *= -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  x *= first;
}
template <typename T, typename... Args>
void read(T &first, Args &...args) {
  read(first);
  read(args...);
}
int T, n, x;
int a[N];
bool vis[N];
int main() {
  read(T);
  while (T--) {
    memset(vis, 0, sizeof(vis));
    int l = 1, r = 2;
    bool poi = 0;
    read(n);
    for (int i = (1); i < (n + 1); ++i) read(a[i]);
    while (r <= n) {
      a[l] < a[r] ? poi = 0 : poi = 1;
      while (r <= n && a[r - 1] < a[r] && !poi) r++;
      while (r <= n && a[r - 1] > a[r] && poi) r++;
      vis[l] = vis[r - 1] = true;
      l = r - 1;
    }
    int num = 0;
    for (int i = (1); i < (n + 1); ++i)
      if (vis[i]) num++;
    cout << num << endl;
    for (int i = (1); i < (n + 1); ++i)
      if (vis[i]) cout << a[i] << " ";
    cout << endl;
  }
  return 0;
}
