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>
const long long maxn = 200010;
long long T, n, top, pla, l, r, cnt;
long long a[maxn], s[maxn], ans[maxn];
inline long long read() {
  char c = getchar();
  long long f = 1, x = 0;
  for (; !isdigit(c); (c == '-') && (f = -1), c = getchar())
    ;
  for (; isdigit(c); x = (x << 1) + (x << 3) + (c ^ 48), c = getchar())
    ;
  return f * x;
}
int main(void) {
  T = read();
  for (; T; --T) {
    n = read();
    top = pla = cnt = 0, l = 1, r = 0;
    for (register int i(1); i <= n; a[i] = read(), ++i)
      ;
    for (register int i(1); i <= n; ++i) {
      if (a[i] > a[i - 1]) {
        if (!pla)
          r = i;
        else {
          pla ^= 1;
          if (l != r)
            s[++top] = l, s[++top] = r;
          else
            s[++top] = r;
          l = i - 1, r = l + 1;
        }
      }
      if (a[i] < a[i - 1]) {
        if (pla)
          r = i;
        else {
          pla ^= 1;
          if (l != r)
            s[++top] = l, s[++top] = r;
          else
            s[++top] = r;
          l = i - 1, r = l + 1;
        }
      }
    }
    s[++top] = l, s[++top] = r;
    for (register int i(1); i <= top; ++i)
      if (a[s[i]] != a[s[i - 1]]) ans[++cnt] = a[s[i]];
    printf("%lld\n", cnt);
    for (register int i(1); i <= cnt; printf("%lld ", ans[i]), ++i)
      ;
    printf("\n");
  }
  return 0;
}
