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;
void solve() {
  long long n, m;
  cin >> n >> m;
  vector<long long> a(n + 1);
  for (long long i = (1); i <= (n); i++) cin >> a[i];
  map<long long, long long> f, f2;
  for (long long i = (1); i <= (n); i++) {
    long long dt;
    if (a[i] == i)
      dt = 0;
    else if (a[i] > i)
      dt = n - a[i] + i;
    else
      dt = i - a[i];
    f[dt]++;
    f2[a[i] - i]++;
  }
  long long biswap = 0;
  for (auto [u, v] : f2) biswap += min(f2[-u], v);
  biswap /= 2;
  vector<long long> ans;
  for (long long k = (0); k <= (n - 1); k++) {
    long long wrong = n - f[k];
    if (wrong <= m + 1)
      ans.emplace_back(k);
    else if (wrong > n - 2 * m)
      continue;
    else {
      vector<pair<long long, long long>> v;
      for (long long i = (1); i <= (n); i++) {
        long long rsp = ((i <= n - k) ? (i + k) : (i - (n - k)));
        v.emplace_back(rsp, a[i]);
      }
      sort(begin(v), end(v));
      vector<long long> vg(n + 1);
      map<long long, long long> pos;
      for (auto [u, v2] : v) pos[v2] = u, vg[u] = v2;
      long long swaps = 0;
      for (long long i = (1); i <= (n); i++)
        if (pos[i] != i) {
          swaps++;
          swap(vg[pos[i]], vg[i]);
          pos[vg[pos[i]]] = pos[i];
          pos[i] = i;
        }
      if (swaps <= m) ans.emplace_back(k);
    }
  }
  cout << ans.size() << " ";
  for (auto e : ans) cout << e << " ";
  cout << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long T = 1;
  cin >> T;
  while (T--) solve();
  return 0;
}
