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 read() {
  long long x = 0, f = 1;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') f = -1;
  } while (ch < '0' || ch > '9');
  do {
    x = x * 10 + ch - 48;
    ch = getchar();
  } while (ch >= '0' && ch <= '9');
  return x * f;
}
const int MAX = 1e5 + 10;
const long long mod = 1e9 + 7;
long long t, n, x, m;
int a[MAX];
int ans[MAX];
void init() { return; }
int main() {
  t = read();
  while (t--) {
    n = read();
    int k = 0;
    int temp;
    for (int i = 1; i <= n; i++) {
      temp = read();
      a[i] = temp;
    }
    int b = 0, now, op = 0;
    int maxi = 0, minn = MAX;
    ans[++k] = a[1];
    if (a[2] > a[1]) {
      maxi = a[2];
      op = 1;
    } else {
      minn = a[2];
      op = 2;
    }
    for (int i = 3; i <= n; i++) {
      if (op == 1) {
        if (a[i] < a[i - 1]) {
          ans[++k] = a[i - 1];
          minn = a[i - 1];
          op = 2;
        }
      } else {
        if (a[i] > a[i - 1]) {
          ans[++k] = a[i - 1];
          maxi = a[i - 1];
          op = 1;
        }
      }
    }
    printf("%d\n", k + 1);
    for (int i = 1; i <= k; i++) {
      if (i != 1) printf(" ");
      printf("%d", ans[i]);
    }
    printf(" %d\n", a[n]);
  }
  return 0;
}
