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 <bits/stdc++.h>
using namespace std;
const long long INF = 2e9 + 9;
vector<long long> adj[300009];
long long vis[300009];
long long n, m;
void dfs(long long x) {
  vis[x] = 1;
  for (auto e : adj[x]) {
    if (vis[e] == -1) dfs(e);
  }
}
long long calculate(vector<long long> need, long long k) {
  long long have[n + 2];
  for (long long i = 0; i < (n); ++i) {
    vis[i] = -1;
    adj[i].clear();
  }
  for (long long i = 0; i < (n); ++i) {
    have[i] = i - k;
    if (have[i] < 0) have[i] += n;
  }
  for (long long i = 0; i < (n); ++i) {
    if (need[i] != have[i]) {
      adj[need[i]].push_back(have[i]);
      adj[have[i]].push_back(need[i]);
    }
  }
  long long comp = 0;
  for (long long i = 0; i < (n); ++i) {
    if (vis[i] == -1) {
      dfs(i);
      comp++;
    }
  }
  return (long long)(need).size() - comp;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    cin >> n >> m;
    vector<long long> vet(n);
    for (long long i = 0; i < (n); ++i) {
      cin >> vet[i];
      vet[i]--;
    }
    long long cont[n + 2];
    vector<long long> ans;
    memset(cont, 0, sizeof cont);
    for (long long i = 0; i < (n); ++i) {
      long long pos = i - vet[i];
      if (pos < 0) pos += n;
      cont[pos]++;
    }
    for (long long i = 0; i < (n); ++i) {
      if (cont[i] >= (n + 2) / 3 && calculate(vet, i) <= m) {
        ans.push_back(i);
      }
    }
    cout << (long long)(ans).size();
    if ((long long)(ans).size() > 0) {
      for (long long i = 0; i < ((long long)(ans).size()); ++i) {
        cout << " " << ans[i];
      }
    }
    cout << "\n";
  }
  return 0;
}
