#include <bits/stdc++.h>
#pragma GCC optimize("O2")
using namespace std;
using namespace std::chrono;
long long mul(long long a, long long b) { return (a * b) % 1000000007; }
long long modinv(long long a) {
  long long b = 1000000007 - 2;
  long long A = a;
  long long c = 1;
  for (long long i = 0; i < 62; i++) {
    if ((long long)1 << i & (b)) {
      c = mul(c, A);
    }
    A *= A;
    A %= 1000000007;
  }
  return c;
}
long long mul(long long a, long long b, long long c) {
  return mul(mul(a, b), c);
}
long long nCr(long long n, long long r) {
  long long num = 1;
  long long den = 1;
  long long k = max(r, n - r);
  for (long long i = k + 1; i <= n; i++) {
    num = mul(num, i);
  }
  for (long long i = 1; i <= n - k; i++) {
    den = mul(den, i);
  }
  return mul(num, modinv(den));
}
long long gcd(long long x, long long y) {
  if (y > x) {
    return gcd(y, x);
  } else if (y == 0) {
    return x;
  } else {
    return gcd(y, x % y);
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long T;
  cin >> T;
  while (T--) {
    long long n, m;
    cin >> n >> m;
    vector<long long> A(n + 1, 0);
    vector<long long> pos(n + 1, 0);
    vector<long long> correct(n + 1, 0);
    for (long long i = 1; i < n + 1; i++) {
      cin >> A[i];
      pos[A[i]] = i;
    }
    map<long long, long long> mp;
    for (long long i = 1; i < n + 1; i++) {
      mp[A[i] - i]++;
    }
    vector<long long> ans;
    for (long long i = 0; i < n; i++) {
      long long total = mp[-1 * (i)];
      if (i > 0) {
        total += mp[(n - i)];
      }
      if (total >= (n - 2 * m)) {
        for (long long j = 1; j < i + 1; j++) {
          correct[j] = j + n - i;
        }
        for (long long j = i + 1; j < n + 1; j++) {
          correct[j] = j - i;
        }
        vector<bool> visited(n + 1, 0);
        total = 0;
        for (long long i = 1; i < n + 1; i++) {
          long long r = i;
          ;
          long long len = 0;
          while (visited[r] == 0) {
            visited[r] = 1;
            len++;
            r = pos[correct[r]];
          }
          if (len) {
            total += len - 1;
          }
        }
        if (total <= m) {
          ans.push_back(i);
        }
      }
    }
    cout << ans.size() << " ";
    for (long long i = 0; i < ans.size(); i++) {
      cout << ans[i] << " ";
    }
    cout << endl;
    ;
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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("O2")
using namespace std;
using namespace std::chrono;
long long mul(long long a, long long b) { return (a * b) % 1000000007; }
long long modinv(long long a) {
  long long b = 1000000007 - 2;
  long long A = a;
  long long c = 1;
  for (long long i = 0; i < 62; i++) {
    if ((long long)1 << i & (b)) {
      c = mul(c, A);
    }
    A *= A;
    A %= 1000000007;
  }
  return c;
}
long long mul(long long a, long long b, long long c) {
  return mul(mul(a, b), c);
}
long long nCr(long long n, long long r) {
  long long num = 1;
  long long den = 1;
  long long k = max(r, n - r);
  for (long long i = k + 1; i <= n; i++) {
    num = mul(num, i);
  }
  for (long long i = 1; i <= n - k; i++) {
    den = mul(den, i);
  }
  return mul(num, modinv(den));
}
long long gcd(long long x, long long y) {
  if (y > x) {
    return gcd(y, x);
  } else if (y == 0) {
    return x;
  } else {
    return gcd(y, x % y);
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long T;
  cin >> T;
  while (T--) {
    long long n, m;
    cin >> n >> m;
    vector<long long> A(n + 1, 0);
    vector<long long> pos(n + 1, 0);
    vector<long long> correct(n + 1, 0);
    for (long long i = 1; i < n + 1; i++) {
      cin >> A[i];
      pos[A[i]] = i;
    }
    map<long long, long long> mp;
    for (long long i = 1; i < n + 1; i++) {
      mp[A[i] - i]++;
    }
    vector<long long> ans;
    for (long long i = 0; i < n; i++) {
      long long total = mp[-1 * (i)];
      if (i > 0) {
        total += mp[(n - i)];
      }
      if (total >= (n - 2 * m)) {
        for (long long j = 1; j < i + 1; j++) {
          correct[j] = j + n - i;
        }
        for (long long j = i + 1; j < n + 1; j++) {
          correct[j] = j - i;
        }
        vector<bool> visited(n + 1, 0);
        total = 0;
        for (long long i = 1; i < n + 1; i++) {
          long long r = i;
          ;
          long long len = 0;
          while (visited[r] == 0) {
            visited[r] = 1;
            len++;
            r = pos[correct[r]];
          }
          if (len) {
            total += len - 1;
          }
        }
        if (total <= m) {
          ans.push_back(i);
        }
      }
    }
    cout << ans.size() << " ";
    for (long long i = 0; i < ans.size(); i++) {
      cout << ans[i] << " ";
    }
    cout << endl;
    ;
  }
  return 0;
}

```