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...
The above is tricky. Write me a correct solution in PYTHON3.

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Can you provide a Python program to find all possible values of k in the cyclic shift operation?
assistant
#Here is Python program to find all possible values of k in the cyclic shift operation.

import sys
input = sys.stdin.readline
from collections import Counter

class DSU:
    def __init__(self, n):
        self.uf = {i: i for i in range(n)}
        self.rank = {i: 1 for i in range(n)}
        self.cnt = n

    def find(self, x):
        if self.uf[x] != x:
            self.uf[x] = self.find(self.uf[x])
        return self.uf[x]

    def union(self, x, y):
        px, py = self.find(x), self.find(y)
        if px != py:
            if self.rank[px] > self.rank[py]:
                px, py = py, px
            self.rank[py] += self.rank[px]
            self.uf[px] = py
            self.cnt -= 1
            return True
        return False

for _ in range(int(input())):
    n, m = map(int, input().split())
    A = list(map(int, input().split()))
    A = [a - 1 for a in A]
    B = [(i - a) % n for i, a in enumerate(A)]
    cnt = Counter(B)
    ans = []
    X = list(range(n))
    for k in range(n):
        if n - cnt[k] > 2 * m: continue
        seen = [0] * n
        cycle = 0
        cur = X[-k:] + X[: -k]
        dsu = DSU(n)
        for a, b in zip(A, cur):
            dsu.union(a, b)
        if n - dsu.cnt <= m:
            ans.append(k)
    print(len(ans), *ans)