human
Can you provide a Python code to find all possible values of k in the cyclic shift operation given the following:
Can you solve the below in PYTHON3?
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.

def naiveSolve():
    
    
    
    return

class UnionFind:
    def __init__(self, n):
        self.parent = list(range(n))
    def find(self, a): #return parent of a. a and b are in same set if they have same parent
        acopy = a
        while a != self.parent[a]:
            a = self.parent[a]
        while acopy != a: #path compression
            self.parent[acopy], acopy = a, self.parent[acopy]
        return a
    def union(self, a, b): #union a and b
        self.parent[self.find(b)] = self.find(a)

def main():
    
    t=int(input())
    allans=[]
    for _ in range(t):
        n,m=readIntArr()
        p=readIntArr()
        
        for i in range(n):
            p[i]-=1 # 0-index
        
        # m<=n//3
        # Best-case, if m==n//3, we can fix at most 2n//3 elements.
        # So, at least n//3 elements must be correct.
        # Need to consider at most 3 permutations.
        
        kMatches=[0]*n
        for i in range(n):
            kMatches[(i-p[i]+n)%n]+=1
        
        matches=[]
        minMatchCnt=n-2*m
        origPerm=list(range(n))
        for k in range(n):
            matchCnts=kMatches[k]
            if matchCnts>=minMatchCnt: # check
                arr=origPerm[n-k:]+origPerm[:n-k]
                
                targetMap=dict() # {value:index}
                for i in range(n):
                    targetMap[arr[i]]=i
                adj=[-1]*n # adj[i] is i's next destination
                for i in range(n):
                    adj[i]=targetMap[p[i]]
                
                # count number of cycles which is number of groups
                uf=UnionFind(n)
                visited=[False]*n
                groups=0
                for i in range(n):
                    if i==adj[i]: continue # already matching
                    if visited[i]: continue
                    groups+=1
                    visited[i]=True
                    node=adj[i]
                    while uf.find(i)!=uf.find(node):
                        visited[node]=True
        ...