#include <bits/stdc++.h>
using namespace std;
const long long maxn = 3e5 + 10;
const long long mod = 1e9 + 7;
const long long m2 = 1e9 + 7;
const long long INF64 = long long(1e18);
const long long max2 = 1e3 + 10;
const long long N = 1000001;
const long long MAXN = 2e5 + 10;
const long long ALPHABET_SIZE = 2;
int cnt = 0;
int vis[maxn];
vector<int> adj[maxn];
vector<int> cyclic_shift(vector<int> v, int k) {
  int ind = (k);
  int n = v.size();
  vector<int> v2(n);
  int i1 = k;
  int i;
  for (i = 0; i < n; ++i, ++i1) {
    v2[i1 % n] = v[i];
  }
  return (v2);
}
void dfs(int p) {
  vis[p] = 1;
  ++cnt;
  for (auto x : adj[p]) {
    if (vis[x] == 0) {
      dfs(x);
    }
  }
}
int find(vector<int> v, vector<int> v2) {
  int i;
  int n = v.size();
  for (i = 0; i < n; ++i) {
    adj[i].clear();
    vis[i] = 0;
  }
  for (i = 0; i < n; ++i) {
    adj[v[i]].push_back(v2[i]);
    adj[v2[i]].push_back(v[i]);
  }
  int ans = 0;
  for (i = 0; i < n; ++i) {
    cnt = 0;
    if (vis[i] == 0) {
      dfs(i);
      ans += (cnt - 1);
    }
  }
  return (ans);
}
vector<vector<int>> vf2;
void printArr(vector<int>& a, int n) {
  vector<int> v;
  for (int i = 0; i < n; i++) v.push_back(a[i]);
  vf2.push_back(v);
}
void heapPermutation(vector<int>& a, int size, int n) {
  if (size == 1) {
    printArr(a, n);
    return;
  }
  for (int i = 0; i < size; i++) {
    heapPermutation(a, size - 1, n);
    if (size % 2 == 1)
      swap(a[0], a[size - 1]);
    else
      swap(a[i], a[size - 1]);
  }
}
int main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(0);
  cout.tie(0);
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    vector<int> v(n);
    int i;
    int ind[n];
    for (i = 0; i < n; ++i) {
      cin >> v[i];
      --v[i];
      ind[v[i]] = i;
    }
    vector<int> vf(n);
    for (i = 0; i < n; ++i) {
      vf[i] = i;
    }
    int ans[n];
    memset(ans, 0, sizeof(ans));
    for (i = 0; i < n; ++i) {
      int x = ind[i] - i;
      if (x < 0) {
        x += n;
      }
      ans[x]++;
    }
    vector<int> ans2;
    for (i = 0; i < n; ++i) {
      if (n - ans[i] <= 2 * k) {
        vector<int> v2 = cyclic_shift(vf, i);
        int x = find(v, v2);
        if (x <= k) {
          ans2.push_back(i);
        }
      }
    }
    cout << ans2.size() << " ";
    for (auto x : ans2) cout << x << " ";
    cout << "\n";
  }
  return (0);
}

### Prompt

Create a solution in Cpp for 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>
using namespace std;
const long long maxn = 3e5 + 10;
const long long mod = 1e9 + 7;
const long long m2 = 1e9 + 7;
const long long INF64 = long long(1e18);
const long long max2 = 1e3 + 10;
const long long N = 1000001;
const long long MAXN = 2e5 + 10;
const long long ALPHABET_SIZE = 2;
int cnt = 0;
int vis[maxn];
vector<int> adj[maxn];
vector<int> cyclic_shift(vector<int> v, int k) {
  int ind = (k);
  int n = v.size();
  vector<int> v2(n);
  int i1 = k;
  int i;
  for (i = 0; i < n; ++i, ++i1) {
    v2[i1 % n] = v[i];
  }
  return (v2);
}
void dfs(int p) {
  vis[p] = 1;
  ++cnt;
  for (auto x : adj[p]) {
    if (vis[x] == 0) {
      dfs(x);
    }
  }
}
int find(vector<int> v, vector<int> v2) {
  int i;
  int n = v.size();
  for (i = 0; i < n; ++i) {
    adj[i].clear();
    vis[i] = 0;
  }
  for (i = 0; i < n; ++i) {
    adj[v[i]].push_back(v2[i]);
    adj[v2[i]].push_back(v[i]);
  }
  int ans = 0;
  for (i = 0; i < n; ++i) {
    cnt = 0;
    if (vis[i] == 0) {
      dfs(i);
      ans += (cnt - 1);
    }
  }
  return (ans);
}
vector<vector<int>> vf2;
void printArr(vector<int>& a, int n) {
  vector<int> v;
  for (int i = 0; i < n; i++) v.push_back(a[i]);
  vf2.push_back(v);
}
void heapPermutation(vector<int>& a, int size, int n) {
  if (size == 1) {
    printArr(a, n);
    return;
  }
  for (int i = 0; i < size; i++) {
    heapPermutation(a, size - 1, n);
    if (size % 2 == 1)
      swap(a[0], a[size - 1]);
    else
      swap(a[i], a[size - 1]);
  }
}
int main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(0);
  cout.tie(0);
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    vector<int> v(n);
    int i;
    int ind[n];
    for (i = 0; i < n; ++i) {
      cin >> v[i];
      --v[i];
      ind[v[i]] = i;
    }
    vector<int> vf(n);
    for (i = 0; i < n; ++i) {
      vf[i] = i;
    }
    int ans[n];
    memset(ans, 0, sizeof(ans));
    for (i = 0; i < n; ++i) {
      int x = ind[i] - i;
      if (x < 0) {
        x += n;
      }
      ans[x]++;
    }
    vector<int> ans2;
    for (i = 0; i < n; ++i) {
      if (n - ans[i] <= 2 * k) {
        vector<int> v2 = cyclic_shift(vf, i);
        int x = find(v, v2);
        if (x <= k) {
          ans2.push_back(i);
        }
      }
    }
    cout << ans2.size() << " ";
    for (auto x : ans2) cout << x << " ";
    cout << "\n";
  }
  return (0);
}

```