#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
using ll = long long;
using vi = vector<int>;
using pi = pair<int, int>;
using vpi = vector<pair<int, int>>;
using pl = pair<ll, ll>;
using vl = vector<ll>;
using vpl = vector<pl>;
using ld = long double;
template <typename T>
using pqg = priority_queue<T, vector<T>, greater<T>>;
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <typename T_container, typename T = typename enable_if<
                                    !is_same<T_container, string>::value,
                                    typename T_container::value_type>::type>
ostream &operator<<(ostream &os, const T_container &v) {
  os << '{';
  string sep;
  for (const T &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
void setIO(string s) {
  freopen((s + ".in").c_str(), "r", stdin);
  freopen((s + ".out").c_str(), "w", stdout);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int INF = 1e9;
const ll LINF = 1e18;
const int MOD = 1e9 + 7;
int n, m;
void dfs(int v, int &cnt, vi &g, vector<bool> &vis) {
  vis[v] = 1;
  cnt++;
  if (!vis[g[v]]) dfs(g[v], cnt, g, vis);
}
int calc(vi &a, vi &perm, vi &pos) {
  vi g(n);
  vector<bool> vis(n, false);
  int res = 0;
  for (int i = 0; i < n; ++i) {
    g[pos[perm[i]]] = (i);
  }
  for (int i = 0; i < n; ++i) {
    if (!vis[i]) {
      dfs(i, res, g, vis);
      res--;
    }
  };
  return res;
}
void solve() {
  cin >> n >> m;
  vi a(n), cnt(n), pos(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    a[i]--;
    pos[a[i]] = i;
    cnt[(i - a[i] + n) % n]++;
  }
  vi ans;
  for (int i = 0; i < n; ++i) {
    if (cnt[i] < n / 3) continue;
    vi perm(n);
    for (int j = 0; j < n; ++j) {
      perm[j] = (j - i + n) % n;
    }
    int v = calc(a, perm, pos);
    if (v <= m) ans.push_back(i);
  }
  cout << (int)(ans).size() << " ";
  for (int x : ans) cout << x << " ";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int testcase;
  cin >> testcase;
  while (testcase--) {
    solve();
  }
}

### Prompt

Please formulate a cpp solution to 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>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
using ll = long long;
using vi = vector<int>;
using pi = pair<int, int>;
using vpi = vector<pair<int, int>>;
using pl = pair<ll, ll>;
using vl = vector<ll>;
using vpl = vector<pl>;
using ld = long double;
template <typename T>
using pqg = priority_queue<T, vector<T>, greater<T>>;
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <typename T_container, typename T = typename enable_if<
                                    !is_same<T_container, string>::value,
                                    typename T_container::value_type>::type>
ostream &operator<<(ostream &os, const T_container &v) {
  os << '{';
  string sep;
  for (const T &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
void setIO(string s) {
  freopen((s + ".in").c_str(), "r", stdin);
  freopen((s + ".out").c_str(), "w", stdout);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int INF = 1e9;
const ll LINF = 1e18;
const int MOD = 1e9 + 7;
int n, m;
void dfs(int v, int &cnt, vi &g, vector<bool> &vis) {
  vis[v] = 1;
  cnt++;
  if (!vis[g[v]]) dfs(g[v], cnt, g, vis);
}
int calc(vi &a, vi &perm, vi &pos) {
  vi g(n);
  vector<bool> vis(n, false);
  int res = 0;
  for (int i = 0; i < n; ++i) {
    g[pos[perm[i]]] = (i);
  }
  for (int i = 0; i < n; ++i) {
    if (!vis[i]) {
      dfs(i, res, g, vis);
      res--;
    }
  };
  return res;
}
void solve() {
  cin >> n >> m;
  vi a(n), cnt(n), pos(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    a[i]--;
    pos[a[i]] = i;
    cnt[(i - a[i] + n) % n]++;
  }
  vi ans;
  for (int i = 0; i < n; ++i) {
    if (cnt[i] < n / 3) continue;
    vi perm(n);
    for (int j = 0; j < n; ++j) {
      perm[j] = (j - i + n) % n;
    }
    int v = calc(a, perm, pos);
    if (v <= m) ans.push_back(i);
  }
  cout << (int)(ans).size() << " ";
  for (int x : ans) cout << x << " ";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int testcase;
  cin >> testcase;
  while (testcase--) {
    solve();
  }
}

```