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>
#pragma GCC optimize( \
    "Ofast,no-stack-protector,unroll-loops,no-stack-protector,fast-math")
using namespace std;
const int N = 1e5 + 5, M = 2 * N + 5;
int n, a[N];
int out[N], sz;
void check() {
  if (sz < 3) return;
  if (out[sz - 1] > out[sz - 2] && out[sz - 2] > out[sz - 3])
    return out[sz - 2] = out[sz - 1], sz--, void();
  if (out[sz - 1] < out[sz - 2] && out[sz - 2] < out[sz - 3])
    return out[sz - 2] = out[sz - 1], sz--, void();
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    for (int i = 0; i < n; ++i) scanf("%d", a + i);
    sz = 0;
    for (int i = 0; i < n;) {
      int j = i + 1;
      for (; j < n && a[j] > a[j - 1]; ++j)
        ;
      int k = i + 1;
      for (; k < n && a[k] < a[k - 1]; ++k)
        ;
      int mx = max(j, k);
      out[sz++] = a[i];
      if (i != mx - 1) out[sz++] = a[mx - 1];
      i = mx;
      check();
    }
    printf("%d\n", sz);
    for (int i = 0; i < sz; ++i) printf("%d%c", out[i], " \n"[i + 1 == sz]);
  }
}
