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;
void _r(const int& x) { scanf("%d", &x); }
void _r(const long long& x) { scanf("%I64d", &x); }
void _r(const char* x) { scanf("%s", x); }
void R() {}
template <class T, class... U>
void R(const T& head, const U&... tail) {
  _r(head);
  R(tail...);
}
void _w(const char x) { putchar(x); }
void _w(const char* x) { printf("%s", x); }
void _w(const int x) { printf("%d", x); }
void _w(const long long x) { printf("%I64d", x); }
void _w(const double x) { printf("%.6f", x); }
void W() {}
template <class T, class... U>
void W(const T& head, const U&... tail) {
  _w(head);
  putchar(sizeof...(tail) ? ' ' : '\n');
  W(tail...);
}
const long long mod = 1e9 + 7;
long long fp1(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans *= a;
    b >>= 1;
    a *= a;
  }
  return ans;
}
long long fp2(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    b >>= 1;
    a = a * a % mod;
  }
  return ans;
}
long long Read() {
  long long res = 0, flag = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    res = res * 10 + ch - '0';
    ch = getchar();
  }
  return res * flag;
}
const int N = 2e5 + 5;
vector<int> v;
int a[N];
int main() {
  int __T;
  scanf("%d", &__T);
  for (int ica = (1); ica <= (__T); ica++) {
    v.clear();
    int n;
    R(n);
    for (int i = (1); i <= (n); i++) R(a[i]);
    v.push_back(a[1]);
    for (int i = (2); i <= (n - 1); i++)
      if (1LL * (a[i] - v[v.size() - 1]) * (a[i + 1] - a[i]) < 0)
        v.push_back(a[i]);
    v.push_back(a[n]);
    cout << v.size() << endl;
    for (auto a : v) {
      cout << a << " ";
    }
    W("");
  }
  return 0;
}
