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 dx[4] = {0, 0, 1, -1};
int dy[4] = {1, -1, 0, 0};
const int mod = (int)1e9 + 7;
const long long INF64 = 3e18;
void smxl(long long &a, long long b) {
  if (a < b) a = b;
}
void smnl(long long &a, long long b) {
  if (a > b) a = b;
}
void adsl(long long &a, long long b) {
  a += b;
  if (a >= mod) a -= mod;
}
void misl(long long &a, long long b) {
  a -= b;
  if (a >= mod) a -= mod;
  if (a < 0) a += mod;
}
void smx(long long &a, long long b) {
  if (a < b) a = b;
}
void smn(long long &a, long long b) {
  if (a > b) a = b;
}
void ads(long long &a, long long b) {
  a += b;
  if (a >= mod) a -= mod;
}
void mis(long long &a, long long b) {
  a -= b;
  if (a >= mod) a -= mod;
  if (a < 0) a += mod;
}
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
long long egcd(long long a, long long b, long long &x, long long &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  long long d = egcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
long long mbinp(long long a, long long b) {
  a %= mod;
  if (b == 0) return 1;
  long long ans = mbinp(a, b / 2);
  long long tmp = (ans * ans) % mod;
  if (b % 2) return ((tmp * a) % mod);
  return ((tmp) % mod);
}
long long binp(long long a, long long b) {
  if (b == 0) return 1;
  long long ans = binp(a, b / 2);
  long long tmp = (ans * ans);
  if (b % 2) return ((tmp * a));
  return ((tmp));
}
long long C(int n, int m) {
  long long ret = 1;
  for (int i = 1; i <= m; i++) {
    ret *= (n - i + 1);
    ret /= i;
  }
  return ret;
}
long long overbinp(long long a, int b) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      if (res < INF64 / a)
        res *= a;
      else
        return INF64;
    }
    if (b > 1) {
      if (a < INF64 / a)
        a *= a;
      else
        return INF64;
    }
    b >>= 1;
  }
  return res;
}
void solve() {
  int n;
  cin >> n;
  vector<int> all(n);
  vector<int> b(n, 0);
  for (int i = 0; i < n; i++) cin >> all[i];
  for (int i = 1; i < n; i++) {
    if (all[i] > all[i - 1])
      b[i] = 1;
    else
      b[i] = 0;
  }
  vector<int> ans;
  ans.push_back(all[0]);
  for (int i = 1; i < n - 1; i++)
    if (b[i] != b[i + 1]) ans.push_back(all[i]);
  ans.push_back(all[n - 1]);
  cout << ans.size() << '\n';
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) solve();
}