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>
namespace io {
template <typename T>
inline void read(T& _x) {
  int f = 0, ch;
  _x = 0;
  while (!isdigit(ch = getchar())) f |= ch == '-';
  while (isdigit(ch)) _x = _x * 10 + ch - '0', ch = getchar();
  if (f) _x = -_x;
}
template <typename T, typename... Args>
inline void read(T& _f, Args&... args) {
  read(_f), read(args...);
}
inline void _deal(char ch) { putchar(ch); }
template <typename T>
inline void _deal(T _x) {
  if (_x < 0) putchar('-'), _x = -_x;
  if (_x > 9) _deal(_x / 10);
  putchar(_x % 10 + '0');
}
inline void write() {}
template <typename T, typename... Args>
inline void write(T _f, Args... args) {
  _deal(_f), write(args...);
}
}  // namespace io
const int N = 1e5 + 5;
int n, p[N], vis[N];
std::vector<int> ans;
int main() {
  int __tests;
  scanf("%d", &__tests);
  while (__tests--) {
    io::read(n), ans.clear();
    for (int i = 1; i <= n; ++i) {
      io::read(p[i]);
      vis[i] = 0;
    }
    if (n == 2) {
      printf("2\n%d %d\n", p[1], p[2]);
      continue;
    }
    ans.push_back(p[1]);
    for (int i = 2; i + 1 <= n; ++i) {
      if (p[i] > ans.back() && p[i] > p[i + 1])
        ans.push_back(p[i]);
      else if (p[i] < ans.back() && p[i] < p[i + 1])
        ans.push_back(p[i]);
    }
    ans.push_back(p[n]);
    printf("%d\n", ans.size());
    for (int i = 0; i < ans.size(); ++i) printf("%d ", ans[i]);
    puts("");
  }
  return 0;
}