description: 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]. 
solution: #include <bits/stdc++.h>
  using namespace std;
  using ll = long long;
  using pii = pair<int, int>;
  using pll = pair<ll, ll>;
  using vi = vector<int>;
  using vl = vector<ll>;
  using vii = vector<pii>;
  using vll = vector<pll>;
  ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
  ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
  vector<int> sieve;
  void calc_sieve(int n) {
    sieve.reserve(n + 1);
    for (int i = 0, nnnn = (n + 1); i < nnnn; ++i) sieve[i] = i;
    for (int i = 2, nnnn = (n + 1); i < nnnn; ++i) {
      if (sieve[i] == i) {
        for (ll j = i * i; j <= n; j += i) {
          if (sieve[j] == j) sieve[j] = i;
        }
      }
    }
  }
  ll norm(ll a, ll P) {
    while (a >= P) a -= P;
    while (a < 0) a += P;
    return a;
  }
  ll mul_mod(ll a, ll b, ll P) { return a * b % P; }
  ll pow_mod(ll a, ll b, ll P) {
    ll curr = a, ans = 1;
    while (b != 0) {
      if (b & 1) {
        ans *= curr;
        ans %= P;
      }
      curr *= curr;
      curr %= P;
      b >>= 1;
    }
    return ans;
  }
  ll inv_mod(ll a, ll P) { return pow_mod(a, P - 2ll, P); }
  ll div_mod(ll a, ll b, ll P) { return a * inv_mod(b, P) % P; }
  int msb(ll a) {
    for (int i = 62; i >= 0; i--) {
      if (a & ((ll)1 << i)) return i + 1;
    }
    return 0;
  }
  vector<ll> fact;
  void calc_fact(int n, ll P) {
    if (fact.size() < 1) fact.push_back(1);
    for (int i = fact.size(), nnnn = (n + 1); i < nnnn; ++i)
      fact.push_back(fact[i - 1] * i % P);
  }
  vector<ll> fact_inv;
  void calc_fact_inv(int n, ll P) {
    if (fact_inv.size() < 1) fact_inv.push_back(1);
    for (int i = fact_inv.size(), nnnn = (n + 1); i < nnnn; ++i)
      fact_inv.push_back(fact_inv[i - 1] * inv_mod(i, P) % P);
  }
  ll nCr(int n, int r, ll P) {
    if (r > n) return 0;
    calc_fact(n, P);
    calc_fact_inv(n, P);
    return (fact[n] * fact_inv[r] % P) * fact_inv[n - r] % P;
  }
  ll nCr_inv(int n, int r, ll P) {
    if (r > n) return 0;
    calc_fact(n, P);
    calc_fact_inv(n, P);
    return (fact_inv[n] * fact[r] % P) * fact[n - r] % P;
  }
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  const long double eps = 1e-9;
  const long long mod = 1000000007;
  const int MAXN = 100000;
  void solve() {
    int n, m;
    cin >> n >> m;
    int p[n], diff[n];
    for (int i = 0, nnnn = (n); i < nnnn; ++i) {
      cin >> p[i], diff[i] = p[i] - i - 1;
      if (diff[i] < 0) diff[i] += n;
    }
    int cnt[n];
    memset(cnt, 0, sizeof cnt);
    for (int i = 0, nnnn = (n); i < nnnn; ++i) cnt[diff[i]]++;
    vector<int> ans;
    for (int i = 0, nnnn = (n); i < nnnn; ++i) {
      if (n - cnt[i] <= m * 2) {
        int diff1[n], count = 0;
        for (int j = 0, nnnn = (n); j < nnnn; ++j) diff1[j] = diff[j];
        int j = 0;
        while (j < n) {
          if (diff1[j] % n != i) {
            int pos = (j + n + diff1[j] - i) % n;
            diff1[j] = (diff1[pos] + n - j + pos) % n;
            diff1[pos] = i;
            count++;
          }
          if (diff1[j] == i) j++;
        }
        if (count <= m) ans.push_back((n - i) % n);
      }
    }
    cout << ans.size() << endl;
    sort((ans).begin(), (ans).end());
    for (int i = 0, nnnn = (ans.size()); i < nnnn; ++i) cout << ans[i] << ' ';
    cout << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout << setprecision(12) << fixed;
    int t;
    cin >> t;
    while (t--) solve();
  }
  