#include <bits/stdc++.h>
using namespace std;
namespace start {
const int inf = INT_MAX >> 1;
long long read() {
  long long x;
  if (scanf("%lld", &x) != 1) exit(0);
  return x;
}
void print(long long x, bool e = 0) { printf("%lld%c", x, " \n"[e]); }
const int N = 500010;
}  // namespace start
using namespace start;
int Count(int a[], int n) {
  static bool vis[N];
  int cnt = 0;
  fill(vis, vis + n, 0);
  for (int i = (0), ib = (n); i < ib; i++)
    if (!vis[i]) {
      for (int p = a[i]; p != i; p = a[p]) vis[p] = 1;
      cnt++;
    }
  return cnt;
}
vector<int> ans;
int a[N], cnt[N];
int b[N];
void Solve() {
  ans.clear();
  int n = read(), k = read();
  fill(cnt, cnt + n, 0);
  for (int i = (0), ib = (n); i < ib; i++) {
    a[i] = read() - 1;
    cnt[(i + n - a[i]) % n]++;
  }
  for (int i = (0), ib = (n); i < ib; i++) {
    if (cnt[i] >= n / 3) {
      for (int j = (0), ib = (n); j < ib; j++) b[j] = a[(j + i) % n];
      if (n - Count(b, n) <= k) ans.push_back(i);
    }
  }
  sort(ans.begin(), ans.end());
  print(ans.size());
  for (int i = (0), ib = (ans.size()); i < ib; i++) print(ans[i], i == ib - 1);
}
signed main() {
  int T = 1;
  T = read();
  for (int ca = (1), ib = (T + 1); ca < ib; ca++) {
    Solve();
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
namespace start {
const int inf = INT_MAX >> 1;
long long read() {
  long long x;
  if (scanf("%lld", &x) != 1) exit(0);
  return x;
}
void print(long long x, bool e = 0) { printf("%lld%c", x, " \n"[e]); }
const int N = 500010;
}  // namespace start
using namespace start;
int Count(int a[], int n) {
  static bool vis[N];
  int cnt = 0;
  fill(vis, vis + n, 0);
  for (int i = (0), ib = (n); i < ib; i++)
    if (!vis[i]) {
      for (int p = a[i]; p != i; p = a[p]) vis[p] = 1;
      cnt++;
    }
  return cnt;
}
vector<int> ans;
int a[N], cnt[N];
int b[N];
void Solve() {
  ans.clear();
  int n = read(), k = read();
  fill(cnt, cnt + n, 0);
  for (int i = (0), ib = (n); i < ib; i++) {
    a[i] = read() - 1;
    cnt[(i + n - a[i]) % n]++;
  }
  for (int i = (0), ib = (n); i < ib; i++) {
    if (cnt[i] >= n / 3) {
      for (int j = (0), ib = (n); j < ib; j++) b[j] = a[(j + i) % n];
      if (n - Count(b, n) <= k) ans.push_back(i);
    }
  }
  sort(ans.begin(), ans.end());
  print(ans.size());
  for (int i = (0), ib = (ans.size()); i < ib; i++) print(ans[i], i == ib - 1);
}
signed main() {
  int T = 1;
  T = read();
  for (int ca = (1), ib = (T + 1); ca < ib; ca++) {
    Solve();
  }
  return 0;
}

```