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...
impor
assistant
#Here is Python code to find all possible values of k in the cyclic shift operation.
t sys
input = sys.stdin.readline

class UnionFindAlgo:

    def __init__(self, n):
        self.par = [i for i in range(n)]

    def find(self, node):
        if self.par[node] == node:
            return node
        self.par[node] = self.find(self.par[node])
        return self.par[node]

    def union(self, u, v):
        a = self.find(u)
        b = self.find(v)
        if a == b:
            return
        self.par[b] = a

t = int(input())
for _ in range(t):
    n, m = map(int, input().split())
    p = list(map(int, input().split()))
    cnt = [0] * n
    for i in range(n):
        cnt[(i+1-p[i]+n) % n] += 1

    ans = []
    for i in range(n):
        if cnt[i]+2*m >= n:
            obj = UnionFindAlgo(n)
            curr = 0
            for j in range(i, n):
                obj.union(curr, p[j]-1)
                curr += 1
            for j in range(0, i):
                obj.union(curr, p[j]-1)
                curr += 1

            connectedRoot = set()
            for j in range(n):
                connectedRoot.add(obj.find(j))
            if n - len(connectedRoot) <= m:
                ans.append(i)
    print(len(ans), *ans)