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]. 

def naiveSolve(n,m,p): # wrong algorithm
    
    ans=[]
    arr=list(range(1,n+1))
    for k in range(n):
        x=0
        for i in range(n):
            if p[i]!=arr[i]:
                x+=1
        if (x+1)//2<=m:
            ans.append(k)
        arr=arr[n-1:]+arr[:n-1]
    return [len(ans)]+ans
    
    return

from collections import Counter

# without 0
# 1
# 6 3
# 2 3 4 5 1 6

# with 0
# 1
# 6 4
# 2 3 4 5 1 6

# with 0
# 1
# 6 3
# 4 3 2 5 1 6

def main():
    
    t=int(input())
    allans=[]
    for _ in range(t):
        n,m=readIntArr()
        p=readIntArr()
        
        rightShiftCnts=Counter() # {rightShift:how many elements}
        for i in range(n):
            desiredPosition=p[i]-1
            rightShift=(i-desiredPosition+n)%n
            rightShiftCnts[rightShift]+=1
        ans=[]
        # print(rightShiftCnts)
        # for k in range(n):
        #     if ((n-rightShiftCnts[k])+1)//2<=m:
        #         ans.append(k)
        
        # every cycle length from 2 increases the required m by 1
        totalAdd=0
        for v in rightShiftCnts.values():
            totalAdd+=max(0,v-1)
        for k in range(n):
            bestMinM=((n-rightShiftCnts[k])+1)//2 # assume all cycle lengths are 1
            currAdd=max(0,rightShiftCnts[k]-1)
            minM=bestMinM+totalAdd-currAdd
            # print('k:{} totalAdd:{} currAdd:{} bestMinM:{} minM:{}'.format(k,totalAdd,currAdd,bestMinM,minM))
            if minM<=m:
                ans.append(k)
        
        ans=[len(ans)]+ans
        allans.append(ans)
        # print(ans)
        # naiveAns=naiveSolve(n,m,p)
        # print(naiveAns)
            
    multiLineArrayOfArraysPrint(allans)
    
    return


import sys
# input=sys.stdin.buffer.readlin,e #FOR READING PURE INTEGER INPUTS (space separation ok)
input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.

def oneLineArrayPrint(arr):
    print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
    print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
    print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
 
def readIntArr():
    return [int(x) for x in input().split()]
# def readFloatArr():
#     return [float(x) for x in input().split()]
 
def makeArr(defaultValFactory,dimensionArr): # eg. makeArr(lambda:0,[n,m])
    dv=defaultValFactory;da=dimensionArr
    if len(da)==1:return [dv() for _ in range(da[0])]
    else:return [makeArr(dv,da[1:]) for _ in range(da[0])]
 
def queryInteractive(l,r):
    print('? {} {}'.format(l,r))
    sys.stdout.flush()
    return int(input())
 
def answerInteractive(x):
    print('! {}'.format(x))
    sys.stdout.flush()
 
inf=float('inf')
# MOD=10**9+7
MOD=998244353

from math import gcd,floor,ceil
# from math import floor,ceil # for Python2
 
for _abc in range(1):
    main()