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]. 
#include<iostream>
#include<algorithm>
#include<cstring>
#define ll long long
using namespace std;
#include<string>
//ll s[100005];
int a[300005];
int fa[300005];
#define inf 999999999
int cnt[300005];
int ans = 0;
int A[4];
int index = 0;
//int sum[6][100005];
//找一个最小的数k 使得n^k>m
int min(int x1, int x2) {
	if (x1 > x2)
		return x2;
	else
		return x1;
}
int max(int x1, int x2) {
	if (x1 < x2)
		return x2;
	else
		return x1;
}

int find(int x) {
	if (fa[x] != x) {
		return fa[x] = find(fa[x]);
	}
	else return fa[x];
}

void init(int n) {
	ans = n;
	memset(A, 0, sizeof(A));
	for(int i=1;i<=n;i++)
		fa[i] = i;
}

void in(int x, int y) {
	int xx = find(x);
	int yy = find(y);
	if (xx != yy) {
		--ans;
		fa[xx] = yy;
	}
}
int main() {
	int t;
	cin >> t;
	while (t--) {
		int n, m;				
		index = 0;
		scanf("%d%d", &n, &m);
		//memset(a, 0, sizeof(a));
		memset(cnt, 0, sizeof(cnt));
		for (int i = 1; i <= n; i++) {
			scanf("%d", &a[i]);
			cnt[(i - a[i]+ n) % n]++;
		}
		for (int k = 0; k <= n - 1; k++) {
			if (n - cnt[k] > m * 2)continue;
			init(n);
			for (int i = 1; i <= n; i++) {
				int aa = (i - k - 1+n) % n + 1;
				in(a[i], aa);
			}
			if (n - ans <= m) {
				A[++index] = k;
			}
		}
		cout << index << " ";
		for (int i = 1; i <= index; i++) {
			cout << A[i] << " ";
		}
		cout << endl;
		
	}
}