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]. 
import sys, math, os
from math import factorial, inf, gcd, pi, sqrt, ceil
from heapq import *
from functools import *
from itertools import *
from collections import *
sys.setrecursionlimit(10**5)

if os.environ.get('HOME') == '/Users/clarity':
    os.chdir(os.path.dirname(os.path.abspath(__file__)))
    inp = open('E_test.txt', 'r').readline

    def debug(*args):
        print(*args, file=sys.stderr)
else:
    inp = sys.stdin.readline

    def debug(*args):
        pass


def line(): return list(map(int, inp().rstrip().split()))
def fline(): return list(map(float, inp().rstrip().split()))
def number(): return int(inp())
def print_array(l): print(' '.join(list(map(str, l))))

def testcase(t):
    for p in range(t):
        solve()

def check_if_shift_is_achievable_in_m_pers(s, shift, m):
    m_left = m
    s = [(x-shift-1)%len(s) for x in s]
    debug(s)
    # check if we can arrange s to match position in m moves.
    to_update = {}
    for i in range(len(s)):
        while s[i] in to_update: 
            s[i] = to_update[s[i]]
        if s[i] != i:
            #put i in s[i] place. But where is i? 
            # remember to update it later
            to_update[i] = s[i]
            s[i] = i
            m_left -=1 
            if m_left < 0:
                return 0
    return 1

            
            

def solve():
    n, m = line()
    s = line()

    dist = [ (s[i] -i -1 + n)%n for i in range(n)]
    c = Counter(dist)
    potential_shifts = [x[0] for x in c.most_common(3)]
    debug(s, dist,  potential_shifts)
    ans = []
    for shift in potential_shifts:
        if check_if_shift_is_achievable_in_m_pers(s, shift, m):
            ans.append((n - shift ) %n )

    print(len(ans),' '.join(list(map(str, sorted(ans)))))

def main():
    testcase(number())

if __name__ == "__main__":
    main()
