#include <bits/stdc++.h>
using namespace std;
vector<long long> vis(3e5 + 10, 0);
vector<long long> temp;
class graph {
 private:
  long long v;
  list<long long>* adj;

 public:
  graph(long long x) {
    v = x + 1;
    adj = new list<long long>[v];
  }
  void adedge(long long a, long long b) {
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  void dfs(long long s);
  void bfs(long long s);
};
void graph::dfs(long long s) {
  vis[s] = 1;
  temp.push_back(s);
  list<long long>::iterator it;
  for (it = adj[s].begin(); it != adj[s].end(); it++) {
    if (!vis[*it]) {
      dfs(*it);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long i, j, k, l, m, n;
    cin >> n >> m;
    long long arr[n + 1];
    long long cnt[n];
    memset(cnt, 0, sizeof cnt);
    for (i = 1; i <= n; i++) {
      cin >> arr[i];
      long long pos = i, num = arr[i];
      if (pos >= num)
        cnt[pos - num] += 1;
      else
        cnt[n - num + pos] += 1;
    }
    vector<long long> ans;
    for (i = 0; i < n; i++) {
      long long a1 = cnt[i];
      long long rem = n - cnt[i];
      if (rem - 1 <= m)
        ans.push_back(i);
      else {
        if (cnt[i] >= n - 2 * m) {
          vector<long long> v2(n + 1, 0);
          vector<long long> v1(n + 1, 0);
          graph g(n);
          for (j = 0; j <= n; j++) vis[j] = 0;
          for (j = 1; j <= n; j++) {
            long long x = n - i + 1;
            if (j <= i)
              v1[j] = n - i + j;
            else
              v1[j] = j - i;
            if (arr[j] > n - i)
              v2[j] = i - (n - arr[j]);
            else
              v2[j] = arr[j] + i;
            g.adedge(j, v2[j]);
          }
          long long tot = 0;
          for (j = 1; j <= n; j++) {
            if (!vis[j]) {
              temp.clear();
              g.dfs(j);
              tot += (long long)temp.size() - 1;
            }
          }
          if (tot <= m) ans.push_back(i);
        }
      }
    }
    cout << (long long)ans.size() << " ";
    for (auto it : ans) cout << it << " ";
    cout << "\n";
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
vector<long long> vis(3e5 + 10, 0);
vector<long long> temp;
class graph {
 private:
  long long v;
  list<long long>* adj;

 public:
  graph(long long x) {
    v = x + 1;
    adj = new list<long long>[v];
  }
  void adedge(long long a, long long b) {
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  void dfs(long long s);
  void bfs(long long s);
};
void graph::dfs(long long s) {
  vis[s] = 1;
  temp.push_back(s);
  list<long long>::iterator it;
  for (it = adj[s].begin(); it != adj[s].end(); it++) {
    if (!vis[*it]) {
      dfs(*it);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long i, j, k, l, m, n;
    cin >> n >> m;
    long long arr[n + 1];
    long long cnt[n];
    memset(cnt, 0, sizeof cnt);
    for (i = 1; i <= n; i++) {
      cin >> arr[i];
      long long pos = i, num = arr[i];
      if (pos >= num)
        cnt[pos - num] += 1;
      else
        cnt[n - num + pos] += 1;
    }
    vector<long long> ans;
    for (i = 0; i < n; i++) {
      long long a1 = cnt[i];
      long long rem = n - cnt[i];
      if (rem - 1 <= m)
        ans.push_back(i);
      else {
        if (cnt[i] >= n - 2 * m) {
          vector<long long> v2(n + 1, 0);
          vector<long long> v1(n + 1, 0);
          graph g(n);
          for (j = 0; j <= n; j++) vis[j] = 0;
          for (j = 1; j <= n; j++) {
            long long x = n - i + 1;
            if (j <= i)
              v1[j] = n - i + j;
            else
              v1[j] = j - i;
            if (arr[j] > n - i)
              v2[j] = i - (n - arr[j]);
            else
              v2[j] = arr[j] + i;
            g.adedge(j, v2[j]);
          }
          long long tot = 0;
          for (j = 1; j <= n; j++) {
            if (!vis[j]) {
              temp.clear();
              g.dfs(j);
              tot += (long long)temp.size() - 1;
            }
          }
          if (tot <= m) ans.push_back(i);
        }
      }
    }
    cout << (long long)ans.size() << " ";
    for (auto it : ans) cout << it << " ";
    cout << "\n";
  }
  return 0;
}

```