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]. 
###pyrival template for fast IO
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
    newlines = 0
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

q=int(input())
from collections import deque
def bfs(graph,currnode,visited):
    stack=deque()
    stack.append(currnode)
    while stack:
        currnode=stack.popleft()
        if visited[currnode]==False:
            visited[currnode]=True

        for neigbour in graph[currnode]:
            if visited[neigbour]==False:
                visited[neigbour]=True
                stack.append(neigbour)
    return ans


def find_minswaps(a,b):
    graph={}
    for i in range(len(a)):
        if a[i]==b[i]:
            graph[a[i]]=[]
        if a[i] not in graph:
            graph[a[i]]=[b[i]]
        else:
            graph[a[i]].append(b[i])
        if b[i] not in graph:
            graph[b[i]]=[a[i]]
        else:
            graph[b[i]].append(a[i])
    visited=[False for x in range(n)]
    count=0
    for i in range(n):
        if visited[i]==False:
            count+=1
            bfs(graph,i,visited)
    return count
while q:
    q-=1
    n,m=[int(x) for x in input().split()]
    arr=[int(x)-1 for x in input().split()]
    ##count number of intergers for each cyclic shift k
    k=[0 for x in range(n)] ##cyclic shift by i
    ##for an i in arr there is only one k such tha arr[i]=(i-k)mod n
    for i in range(n):
        x=arr[i];y=arr[i]-n
        if i-x>=0:
            k[i-x]+=1
        else:
            k[i-y]+=1
    #get the valid k"s using unswaped indices as we have no idea about swaped ones
    tocheck=[]
    ###changed<=2*m; -chnaged>=-2*m
    ##n-changed>=n-2*m
    for i in range(n):
        if k[i]>=n-2*m:
            tocheck.append(i)
    ans=[]
    ##for each k shift permutation find min swaps required to get arr(final arr )
    while tocheck:
        k=tocheck.pop()
        a=[(i-k)%n for i in range(n)]
        ##make the new kshofted arr
        s=n-find_minswaps(a,arr)
        if s<=m:
            ans.append(k)
    ans.sort()
    sys.stdout.write(str(len(ans))+" ")
    for i in range(len(ans)-1):
        sys.stdout.write(str(ans[i])+" ")
    if ans:
        sys.stdout.write(str(ans[-1])+"\n")
    else:
        sys.stdout.write("\n")



