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;
const int N = 2e5 + 5, inf = 0x3f3f3f3f, mod = 1e9 + 7;
int T, n, m, k, x, y, p, t, w, ans, fl, a[N], b[N], c[N], f[N], v[N];
inline long long read() {
  char c = getchar();
  long long a = 0, b = 1;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') b = -1;
  for (; c >= '0' && c <= '9'; c = getchar()) a = a * 10 + c - 48;
  return a * b;
}
int main() {
  T = read();
  while (T--) {
    n = read();
    w = 0, t = 0;
    for (int i = 1; i <= n; i++) v[i] = 0;
    for (int i = 1; i <= n; i++) a[i] = read();
    for (int i = 2; i < n; i++)
      if (a[i - 1] > a[i] && a[i] > a[i + 1]) v[i] = 1;
    for (int i = 1; i <= n; i++)
      if (!v[i]) b[++w] = a[i];
    memset(v, 0, sizeof(v));
    for (int i = 2; i < w; i++)
      if (b[i - 1] < b[i] && b[i] < b[i + 1]) v[i] = 1;
    for (int i = 1; i <= w; i++)
      if (!v[i]) c[++t] = b[i];
    cout << t << endl;
    for (int i = 1; i <= t; i++) {
      cout << c[i] << ' ';
    }
    cout << endl;
  }
  return 0;
}
