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;
template <typename T, typename U>
static inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static inline void amax(T &x, U y) {
  if (x < y) x = y;
}
long long cdiv(long long a, long long b) {
  return a / b + ((a ^ b) > 0 && a % b);
}
long long fdiv(long long a, long long b) {
  return a / b - ((a ^ b) < 0 && a % b);
}
long long max(long long a, long long b, long long c) {
  return max(a, max(b, c));
}
long long min(long long a, long long b, long long c) {
  return min(a, min(b, c));
}
long long gcd(long long a, long long b) {
  if (b == 0)
    return (a);
  else
    return (gcd(b, a % b));
}
long long inverseMod(long long a, long long b) {
  long long r = 1;
  while (b > 0) {
    if (b & 1) r = r * a % 1000000007;
    a = a * a % 1000000007;
    b /= 2;
  }
  return r;
}
long long comb(long long m, long long k) {
  long long r = 1;
  long long d;
  for (d = 1; d <= k; d++) {
    r *= m--;
    r /= d;
  }
  return r;
}
bool prime[10000005];
long long hah[10000005];
void sieve(long long n = 10000002) {
  memset(prime, true, sizeof(prime));
  memset(hah, 0, sizeof(hah));
  for (long long p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      hah[p] = p;
      for (long long i = p * p; i <= n; i += p) {
        prime[i] = false;
        if (hah[i] == 0) hah[i] = p;
      }
    }
  }
}
long long dp[1000005];
void initialize(long long n = 1000000) {
  memset(dp, 0, sizeof(dp));
  dp[0] = 1;
  for (long long i = 1; i < n;) {
    dp[i] = dp[i - 1] + 18;
    i++;
    if (i == 2) {
      for (long long j = 2; j <= 9; j++) {
        dp[i] = dp[i - 1] + 9;
        i++;
      }
    } else {
      for (long long j = 1; j <= 9; j++) {
        dp[i] = dp[i - 1] + 9;
        i++;
      }
    }
  }
}
int _destroyer_() {
  long long n;
  cin >> n;
  long long num[n];
  for (long long i = 0; i < n; i++) cin >> num[i];
  if (n == 2) return cout << 2 << "\n" << num[0] << " " << num[1] << "\n", 0;
  long long dp[n];
  dp[0] = 0;
  vector<long long> v;
  v.push_back(0);
  for (long long i = 1; i < n; i++) {
    dp[i] = num[i] - num[i - 1];
  }
  for (long long i = 2; i < n - 1; i++) {
    if (dp[i] * dp[i - 1] < 0) {
      v.push_back(i - 1);
    }
  }
  if (dp[n - 1] * dp[n - 2] > 0)
    v.push_back(n - 1);
  else {
    v.push_back(n - 2);
    v.push_back(n - 1);
  }
  cout << v.size() << "\n";
  for (auto c : v) cout << num[c] << " ";
  cout << "\n";
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  sieve();
  long long TESTS = 1;
  cin >> TESTS;
  while (TESTS--) _destroyer_();
  return 0;
}