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 array [2, 3, 1, 4]; 
  * in the second test case, the only possible value for the cyclic shift is 0. If we shift [1, 2, 3] by 0 positions, we get [1, 2, 3]. Then we don't change the array at all (we stated that we made at most 1 swap), so the resulting array stays [1, 2, 3]; 
  * in the third test case, all values from 0 to 2 are possible for the cyclic shift: 
    * if we shift [1, 2, 3] by 0 positions, we get [1, 2, 3]. Then we can swap the 1-st and the 3-rd elements to get [3, 2, 1]; 
    * if we shift [1, 2, 3] by 1 position, we get [3, 1, 2]. Then we can swap the 2-nd and the 3-rd elements to get [3, 2, 1]; 
    * if we shift [1, 2, 3] by 2 positions, we get [2, 3, 1]. Then we can swap the 1-st and the 2-nd elements to get [3, 2, 1]; 
  * in the fourth test case, we stated that we didn't do any swaps after the cyclic shift, but no value of cyclic shift could produce the array [1, 2, 3, 4, 6, 5]. 
import io,os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline



def corr(num,k,n):
    if num>k:   return num-k-1
    else:  return num+k-1

 

def examine(arr,k,m):

    n = len(arr)
    visited = [False]*n
    tot = 0
    for i in range(n):
        visited[i] = True
        if corr(arr[i],k,n)==i:  continue 
        j = i
        temp = 0
        while not visited[j]:
            j = corr(arr[j],k,n)
            visited[j] = True
            temp += 1
        tot += temp

    return tot<=m
        

        
             
 
         
    





def main(t):


    n,m = map(int, input().split())
    arr = list(map(int,input().split()))


    ans = 0
    dic = {}
    for i in range(n):
        if arr[i]>=i+1:  diff = arr[i]-i-1
        else:  diff = arr[i]-1-i+n
        if diff not in dic:   dic[diff] = 0
        dic[diff] += 1


#    print(dic)

    ans = []
    for key in dic:
        if dic[key] >= n - 2*m:
            if examine(arr,key,m):
                if key==0:  ans.append(0)
                else:      ans.append(n-key)
         

    ans = sorted(ans)
    ans.insert(0,len(ans))

    ans = " ".join(map(str,ans))
    print(ans)




















T = int(input())
t = 1
while t<=T:
    main(t)
    t += 1
