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;
long long n, a[100005], pre[100005], t[100005];
long long sta[100005], top;
long long f[100005];
void work() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++)
    scanf("%lld", &a[i]), f[i] = t[i] = pre[i] = 0;
  long long xx, yy, tx, ty;
  xx = -a[1];
  yy = a[1];
  tx = ty = 1;
  for (long long i = 2; i <= n; i++) {
    if (a[i] + xx > yy - a[i])
      f[i] = a[i] + xx, pre[i] = tx, t[i] = t[tx] + 1;
    else if (a[i] + xx == yy - a[i]) {
      if (t[tx] < t[ty])
        f[i] = a[i] + xx, pre[i] = tx, t[i] = t[tx] + 1;
      else
        f[i] = yy - a[i], pre[i] = ty, t[i] = t[ty] + 1;
    } else
      f[i] = yy - a[i], pre[i] = ty, t[i] = t[ty] + 1;
    if (f[i] - a[i] > xx)
      xx = f[i] - a[i], tx = i;
    else if (f[i] - a[i] == xx && t[i] < t[tx])
      tx = i;
    if (f[i] + a[i] > yy)
      yy = f[i] + a[i], ty = i;
    else if (f[i] + a[i] == yy && t[i] < t[ty])
      ty = i;
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++)
    if (f[i] > f[ans]) ans = i;
  printf("%lld\n", t[ans] + 1);
  while (ans) sta[++top] = ans, ans = pre[ans];
  while (top) printf("%lld ", a[sta[top--]]);
  putchar('\n');
}
signed main() {
  long long T;
  scanf("%lld", &T);
  while (T--) work();
  return 0;
}
