#include <bits/stdc++.h>
using namespace std;
long long int power(long long int x, unsigned long long int y) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) {
      res = res * x;
    }
    y = y >> 1;
    x = x * x;
  }
  return res;
}
long long int powermod(long long int x, unsigned long long int y,
                       long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) {
      res = (res * x) % p;
    }
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int dfs(vector<long long int> &p, vector<long long int> &v) {
  long long int n = v.size() - 1;
  vector<long long int> vis(n + 1), idx(n + 1);
  for (long long int i = (long long int)(1); i < (long long int)(n + 1); i++) {
    idx[p[i]] = i;
  }
  long long int ans = 0;
  for (long long int i = (long long int)(1); i < (long long int)(n + 1); i++) {
    if (vis[i]) continue;
    long long int c = 0;
    long long int node = i;
    while (!vis[node]) {
      vis[node] = 1;
      long long int go_to = v[node];
      node = idx[go_to];
      c++;
    }
    ans += (c - 1);
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int T;
  cin >> T;
  while (T--) {
    long long int n, m;
    cin >> n >> m;
    vector<long long int> v, f(n + 1), diff[n + 1];
    v.push_back(0);
    for (long long int i = (long long int)(0); i < (long long int)(n); i++) {
      long long int t;
      cin >> t;
      v.push_back(t);
      f[t] = i + 1;
    }
    for (long long int i = (long long int)(1); i < (long long int)(n + 1);
         i++) {
      long long int val = v[i];
      if (i >= v[i]) {
        diff[i - v[i]].push_back(v[i]);
      } else {
        long long int d = (n - v[i]);
        d += i;
        diff[d].push_back(v[i]);
      }
    }
    vector<long long int> ans;
    for (long long int i = (long long int)(0); i < (long long int)(n); i++) {
      long long int val = diff[i].size();
      if (val < n / 3) continue;
      vector<long long int> p(n + 1);
      for (long long int j = (long long int)(0); j < (long long int)(n); j++)
        p[(j + i) % n + 1] = j + 1;
      long long int rem = (n - val);
      if (dfs(p, v) <= m) ans.push_back(i);
    }
    cout << ans.size() << " ";
    for (const auto itr : ans) {
      cout << itr << ' ';
    }
    cout << "\n";
    ;
  }
  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;
long long int power(long long int x, unsigned long long int y) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) {
      res = res * x;
    }
    y = y >> 1;
    x = x * x;
  }
  return res;
}
long long int powermod(long long int x, unsigned long long int y,
                       long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) {
      res = (res * x) % p;
    }
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int dfs(vector<long long int> &p, vector<long long int> &v) {
  long long int n = v.size() - 1;
  vector<long long int> vis(n + 1), idx(n + 1);
  for (long long int i = (long long int)(1); i < (long long int)(n + 1); i++) {
    idx[p[i]] = i;
  }
  long long int ans = 0;
  for (long long int i = (long long int)(1); i < (long long int)(n + 1); i++) {
    if (vis[i]) continue;
    long long int c = 0;
    long long int node = i;
    while (!vis[node]) {
      vis[node] = 1;
      long long int go_to = v[node];
      node = idx[go_to];
      c++;
    }
    ans += (c - 1);
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int T;
  cin >> T;
  while (T--) {
    long long int n, m;
    cin >> n >> m;
    vector<long long int> v, f(n + 1), diff[n + 1];
    v.push_back(0);
    for (long long int i = (long long int)(0); i < (long long int)(n); i++) {
      long long int t;
      cin >> t;
      v.push_back(t);
      f[t] = i + 1;
    }
    for (long long int i = (long long int)(1); i < (long long int)(n + 1);
         i++) {
      long long int val = v[i];
      if (i >= v[i]) {
        diff[i - v[i]].push_back(v[i]);
      } else {
        long long int d = (n - v[i]);
        d += i;
        diff[d].push_back(v[i]);
      }
    }
    vector<long long int> ans;
    for (long long int i = (long long int)(0); i < (long long int)(n); i++) {
      long long int val = diff[i].size();
      if (val < n / 3) continue;
      vector<long long int> p(n + 1);
      for (long long int j = (long long int)(0); j < (long long int)(n); j++)
        p[(j + i) % n + 1] = j + 1;
      long long int rem = (n - val);
      if (dfs(p, v) <= m) ans.push_back(i);
    }
    cout << ans.size() << " ";
    for (const auto itr : ans) {
      cout << itr << ' ';
    }
    cout << "\n";
    ;
  }
  return 0;
}

```