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].
#include <bits/stdc++.h>
using namespace std;
typedef pair<long long, long long> ll;
typedef vector<long long> vl;
typedef vector<ll> vll;
typedef vector<vl> vvl;
long long checkpoint_counter = 0;
template <typename T>
ostream &operator<<(ostream &o, vector<T> v) {
  if (v.size() > 0) o << v[0];
  for (unsigned i = 1; i < v.size(); i++) o << " " << v[i];
  return o << " ";
}
template <typename U, typename V>
ostream &operator<<(ostream &o, pair<U, V> p) {
  return o << "(" << p.first << ", " << p.second << ") ";
}
template <typename T>
istream &operator>>(istream &in, vector<T> &v) {
  for (unsigned i = 0; i < v.size(); i++) in >> v[i];
  return in;
}
template <typename T>
istream &operator>>(istream &in, pair<T, T> &p) {
  in >> p.first;
  in >> p.second;
  return in;
}
template <typename T>
ostream &operator<<(ostream &o, set<T> v) {
  for (auto &it : v) o << it << " ";
  return o << "\n";
}
template <typename T, typename U>
ostream &operator<<(ostream &o, map<T, U> v) {
  for (auto &it : v) o << it << " ";
  return o << "\n";
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
bool cmp(ll a, ll b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first < b.first;
}
long long tree[600100];
void build(long long node, long long start, long long end) {
  if (start == end) {
    tree[node] = 0;
  } else {
    long long mid = (start + end) / 2;
    build(2 * node + 1, start, mid);
    build(2 * node + 2, mid + 1, end);
    tree[node] = tree[2 * node + 1] + tree[2 * node + 2];
  }
}
void update(long long node, long long start, long long end, long long idx,
            long long val) {
  if (start == end) {
    tree[node] = val;
  } else {
    long long mid = (start + end) / 2;
    if (start <= idx and idx <= mid) {
      update(2 * node + 1, start, mid, idx, val);
    } else {
      update(2 * node + 2, mid + 1, end, idx, val);
    }
    tree[node] = max(tree[2 * node + 1], tree[2 * node + 2]);
  }
}
long long query(long long node, long long start, long long end, long long l,
                long long r) {
  if (r < start or end < l) {
    return 0;
  }
  if (l <= start and end <= r) {
    return tree[node];
  }
  long long mid = (start + end) / 2;
  long long p1 = query(2 * node, start, mid, l, r);
  long long p2 = query(2 * node + 1, mid + 1, end, l, r);
  return max(p1, p2);
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, te;
    vl ve;
    cin >> n;
    for (long long i = (0); i < (long long)n; i++) {
      cin >> te;
      ve.push_back((te));
    }
    vl ans;
    long long a = ve[0], b = ve[1], c;
    ans.push_back((a));
    for (long long i = (2); i < (long long)n; i++) {
      c = ve[i];
      if (a < b && b < c) {
        b = c;
      } else if (a > b && b > c) {
        b = c;
      } else {
        ans.push_back((b));
        a = b;
        b = c;
      }
    }
    ans.push_back(b);
    cout << ans.size() << "\n";
    cout << ans << "\n";
  }
  return 0;
}
