human
An identity permutation of length n is an array [1, 2, 3, ..., n].

We performed the following operations to an identity permutation of length n:

  * firstly, we cyclically shifted it to the right by k positions, where k is unknown to you (the only thing you know is that 0 ≤ k ≤ n - 1). When an array is cyclically shifted to the right by k positions, the resulting array is formed by taking k last elements of the original array (without changing their relative order), and then appending n - k first elements to the right of them (without changing relative order of the first n - k elements as well). For example, if we cyclically shift the identity permutation of length 6 by 2 positions, we get the array [5, 6, 1, 2, 3, 4].

  * secondly, we performed the following operation at most m times: pick any two elements of the array and swap them. 


You are given the values of n and m, and the resulting array. Your task is to find all possible values of k in the cyclic shift operation.

Input

The first line contains one integer t (1 ≤ t ≤ 10^5) — the number of test cases.

Each test case consists of two lines. The first line contains two integers n and m (3 ≤ n ≤ 3 ⋅ 10^5.
0 ≤ m ≤ n/3).

The second line contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, each integer from 1 to n appears in this sequence exactly once) — the resulting array.

The sum of n over all test cases does not exceed 3 ⋅ 10^5.

Output

For each test case, print the answer in the following way:

  * firstly, print one integer r (0 ≤ r ≤ n) — the number of possible values of k for the cyclic shift operation.

  * secondly, print r integers k_1, k_2, ..., k_r (0 ≤ k_i ≤ n - 1) — all possible values of k in increasing order. 

Example

Input

4
4 1
2 3 1 4
3 1
1 2 3
3 1
3 2 1
6 0
1 2 3 4 6 5

Output

1 3
1 0
3 0 1 2
0

Note

Consider the example: 

  * in the first test case, the only possible value for the cyclic shift is 3. If we shift [1, 2, 3, 4] by 3 positions, we get [2, 3, 4, 1]. Then we can swap the 3-rd and the 4-th elements to get the...
assistant
#Here is Python code to find all possible values of k in the cyclic shift operation.

import sys

IS_INTERACTIVE = False
input = input  # type: ignore

# input: function
if not IS_INTERACTIVE:
    *data, = sys.stdin.read().split("\n")[::-1]

    def input():  # type: ignore
        return data.pop()

def fprint(*args, **kwargs):
    print(*args, **kwargs, flush=True)

def eprint(*args, **kwargs):
    print(*args, **kwargs, file=sys.stderr)

def check(arr, k, m):
    arr = arr[k:] + arr[:k]
    solved = [0] * len(arr)
    cnt = 0
    for i in range(len(arr)):
        if not solved[i] and i != arr[i]:
            length = 1
            cur = arr[i]
            solved[cur] = 1
            while cur != i:
                length += 1
                cur = arr[cur]
                solved[cur] = 1
            cnt += max(0, length - 1)
    return cnt <= m

supans = []
for i in range(int(input())):
    n, m = map(int, input().split())
    *arr, = map(int, input().split())
    arr = [v - 1 for v in arr]
    shifts = [0] * n
    for i, v in enumerate(arr):
        shifts[(i - v) % n] += 1

    ans = []
    for k in range(n):
        if shifts[k] >= (n - 2 * m):
            if check(arr, k, m):
                ans.append(k)
    fmt = "%s %s" % (len(ans), " ".join(map(str, ans)))
    supans.append(fmt)
print("\n".join(supans))