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;
const int Maxn = 1e6 + 1;
const int Mod = 1e9 + 7;
const long long int MOD = 1000000007LL;
const long long int MAX = 100010LL;
template <typename T>
T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <typename T>
T power(T x, T y, long long int m = MOD) {
  T ans = 1;
  x %= m;
  while (y > 0) {
    if (y & 1ll) ans = (ans * x) % m;
    y >>= 1ll;
    x = (x * x) % m;
  }
  return ans % m;
}
template <typename T, typename T1>
T mod(T x, T1 p) {
  x %= p;
  if (x < 0) x += p;
  return x;
}
void leastPrimeFactor(vector<int>& least_prime, int n) {
  least_prime[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (least_prime[i] == 0) {
      least_prime[i] = i;
      for (int j = 2 * i; j <= n; j += i)
        if (least_prime[j] == 0) least_prime[j] = i;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  for (int ze = 0; ze < t; ze++) {
    int n;
    cin >> n;
    vector<long long int> p(n);
    for (int i = 0; i < n; i++) cin >> p[i];
    vector<int> ans;
    ans.push_back(p[0]);
    int s = -1;
    for (int i = 1; i < n; i++) {
      if (s == -1) {
        if (p[i] > p[i - 1])
          s = 1;
        else if (p[i] < p[i - 1])
          s = 0;
      } else {
        if (s == 1 && p[i] < p[i - 1]) {
          ans.push_back(p[i - 1]);
          i--;
          s = -1;
        }
        if (s == 0 && p[i] > p[i - 1]) {
          ans.push_back(p[i - 1]);
          i--;
          s = -1;
        }
      }
    }
    if (ans.back() != p.back()) ans.push_back(p.back());
    cout << ans.size() << endl;
    for (int i : ans) cout << i << " ";
    cout << endl;
  }
}