#include <bits/stdc++.h>
using namespace std;
const int N = 300005;
int father[N], a[N], b[N];
int getfather(int x) {
  return father[x] == x ? x : father[x] = getfather(father[x]);
}
void init(int n) {
  for (int i = 1; i <= n; ++i) {
    father[i] = i;
  }
}
void solve() {
  int n, m;
  cin >> n >> m;
  map<int, int> mp;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
    if (a[i] > i)
      mp[n - a[i] + i]++;
    else
      mp[i - a[i]]++;
  }
  vector<int> res, tp;
  for (auto i : mp) {
    if (i.second >= n - 2 * m) tp.push_back(i.first);
  }
  sort(tp.begin(), tp.end());
  tp.erase(unique(tp.begin(), tp.end()), tp.end());
  for (auto k : tp) {
    for (int i = 1; i <= n; ++i) {
      if (i <= k)
        b[i] = i + n - k;
      else
        b[i] = i - k;
    }
    init(n);
    for (int j = 1; j <= n; j++) father[getfather(a[j])] = getfather(b[j]);
    map<int, int> c;
    for (int j = 1; j <= n; j++) c[getfather(j)]++;
    int cnt = 0;
    for (auto x : c) cnt += x.second - 1;
    if (cnt <= m) res.push_back(k);
  }
  sort(res.begin(), res.end());
  cout << res.size() << " ";
  for (int i = 0; i < res.size(); ++i) {
    printf("%d ", res[i]);
  }
  printf("\n");
}
signed main() {
  int T;
  scanf("%d", &T);
  while (T--) solve();
}

### Prompt

Please create a solution in Cpp to the following problem:

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]. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 300005;
int father[N], a[N], b[N];
int getfather(int x) {
  return father[x] == x ? x : father[x] = getfather(father[x]);
}
void init(int n) {
  for (int i = 1; i <= n; ++i) {
    father[i] = i;
  }
}
void solve() {
  int n, m;
  cin >> n >> m;
  map<int, int> mp;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
    if (a[i] > i)
      mp[n - a[i] + i]++;
    else
      mp[i - a[i]]++;
  }
  vector<int> res, tp;
  for (auto i : mp) {
    if (i.second >= n - 2 * m) tp.push_back(i.first);
  }
  sort(tp.begin(), tp.end());
  tp.erase(unique(tp.begin(), tp.end()), tp.end());
  for (auto k : tp) {
    for (int i = 1; i <= n; ++i) {
      if (i <= k)
        b[i] = i + n - k;
      else
        b[i] = i - k;
    }
    init(n);
    for (int j = 1; j <= n; j++) father[getfather(a[j])] = getfather(b[j]);
    map<int, int> c;
    for (int j = 1; j <= n; j++) c[getfather(j)]++;
    int cnt = 0;
    for (auto x : c) cnt += x.second - 1;
    if (cnt <= m) res.push_back(k);
  }
  sort(res.begin(), res.end());
  cout << res.size() << " ";
  for (int i = 0; i < res.size(); ++i) {
    printf("%d ", res[i]);
  }
  printf("\n");
}
signed main() {
  int T;
  scanf("%d", &T);
  while (T--) solve();
}

```