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]. 
from collections import Counter, defaultdict, deque
import bisect
from sys import stdin, stdout
from itertools import repeat, permutations
import math
import heapq
 
MOD = 998244353
def inp(force_list=False):
    re = map(int, raw_input().split())
    if len(re) == 1 and not force_list:
        return re[0]
    return re
 
def inpf(force_list=False):
    re = map(float, raw_input().split())
    if len(re) == 1 and not force_list:
        return re[0]
    return re
 
def inst():
    return raw_input().strip()
 
 
def gcd(x, y):
   while(y):
       x, y = y, x % y
   return x
 
def lcm(x, y):
    return x*y/(gcd(x,y))
 
def qmod(a, b, mod=MOD):
    res = 1
    while b:
        if b&1:
            res = (res*a)%mod
        b >>= 1
        a = (a*a)%mod
        # print b
    return res
 
def inv(a):
    return qmod(a, MOD-2)
 
def query(num, k):
    print tmk(num, k)
    stdout.flush()
    return inp()
 
def my_main():
    kase = inp()
    pans = []
    for _ in range(kase):
        n, m = inp()
        da = inp(True)
        ans = []
        diff = [i-da[i]+1 for i in range(n)]
        ct = sorted(Counter(diff).items(), key=lambda v:-v[-1])
        for k, v in ct[:6]:
            ta = [-1] * n
            if k < 0:
                pos = 0
                ta[pos] = -k
            else:
                pos = n-1
                ta[pos] = n-1-k
            last = ta[pos]
            pos = (pos+1)%n
            while ta[pos] == -1:
                last = (last+1)%n
                ta[pos] = last
                pos = (pos+1)%n
            pos = [0] * n
            for i in range(n):
                pos[ta[i]] = i
            flag = [0]*n
            tm = 0
            for i in range(n):
                if flag[i]:
                    continue
                now = pos[da[i]-1]
                flag[i] = 1
                while flag[now]==0:
                    tm += 1
                    flag[now] = 1
                    now = pos[da[now]-1]
            # print pos, ta, da, tm
            if tm <= m:
                ans.append((n-ta[0])%n)
                # print ta, k, tm
        ans.sort()
        ans.insert(0, len(ans))
        pans.append(' '.join(map(str, ans)))
        # print '\n'.join(map(str, ans))


 
    print '\n'.join(pans)

my_main()