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 MOD = 1000000007LL;
long long int fastpow(long long int base, long long int exp) {
  long long int res = 1;
  while (exp > 0) {
    if (exp % 2 == 1) res = (res * base) % MOD;
    base = (base * base) % MOD;
    exp /= 2;
  }
  return res % MOD;
}
template <typename T>
T mult(T x, T y, T mod) {
  return (((x % mod) * (y % mod)) % mod);
}
template <typename T>
T add(T x, T y, T mod) {
  return (((x % mod) + (y % mod)) % mod);
}
template <typename T>
T gcd(T x, T y) {
  return (!y ? x : gcd(y, x % y));
}
template <typename T>
T lcm(T x, T y) {
  return ((x * y) / gcd(x, y));
}
int32_t main() {
  long long int testcases;
  cin >> testcases;
  while (testcases--) {
    long long int n, p[100005];
    cin >> n;
    for (int i = 0; i < n; i++) cin >> p[i];
    vector<long long int> AC;
    for (int i = 0; i < n - 1;) {
      AC.push_back(p[i]);
      if (p[i + 1] > p[i]) {
        while (i + 1 != n && p[i + 1] > p[i]) i++;
      } else if (p[i + 1] <= p[i]) {
        while (i + 1 != n && p[i + 1] <= p[i]) i++;
      }
    }
    if (p[n - 1] != AC[AC.size() - 1]) AC.push_back(p[n - 1]);
    cout << AC.size() << endl;
    long long int len = AC.size();
    for (int i = 0; i < len; i++) cout << AC[i] << " ";
    cout << endl;
  }
  return 0;
}