#include <bits/stdc++.h>
using namespace std;
vector<int> vec;
int p[300005], to[300005], cnt[300005];
int read() {
  int A;
  bool K;
  char C;
  C = A = K = 0;
  while (C < '0' || C > '9') K |= C == '-', C = getchar();
  while (C > '/' && C < ':') A = (A << 3) + (A << 1) + (C ^ 48), C = getchar();
  return (K ? -A : A);
}
void write(int X) {
  if (X < 0) putchar('-'), X = -X;
  if (X > 9) write(X / 10);
  putchar(X % 10 | 48);
}
int main() {
  int t, n, m, s, j, i, h, k;
  t = read();
  while (t--) {
    n = read(), m = read();
    for (i = 1; i <= (n); i++) cnt[i] = 0;
    for (i = 1; i <= (n); i++) p[i] = read(), cnt[(p[i] - i + n) % n]++;
    vec.clear();
    for (k = 0; k <= (n - 1); k++)
      if (cnt[k] >= n - (m << 1)) {
        for (i = 1; i <= (n); i++) to[((i + k) % n ? (i + k) % n : n)] = p[i];
        s = n;
        for (i = 1; i <= (n); i++)
          if (to[i]) {
            j = i;
            while (to[j]) {
              h = to[j];
              to[j] = 0;
              j = h;
            }
            s--;
          }
        if (s <= m) vec.push_back(k);
      }
    write(vec.size());
    if (!vec.empty() && !vec[0]) vec.erase(vec.begin()), vec.push_back(0);
    while (!vec.empty()) {
      putchar(' ');
      write((n - vec.back()) % n);
      vec.pop_back();
    }
    putchar('\n');
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
vector<int> vec;
int p[300005], to[300005], cnt[300005];
int read() {
  int A;
  bool K;
  char C;
  C = A = K = 0;
  while (C < '0' || C > '9') K |= C == '-', C = getchar();
  while (C > '/' && C < ':') A = (A << 3) + (A << 1) + (C ^ 48), C = getchar();
  return (K ? -A : A);
}
void write(int X) {
  if (X < 0) putchar('-'), X = -X;
  if (X > 9) write(X / 10);
  putchar(X % 10 | 48);
}
int main() {
  int t, n, m, s, j, i, h, k;
  t = read();
  while (t--) {
    n = read(), m = read();
    for (i = 1; i <= (n); i++) cnt[i] = 0;
    for (i = 1; i <= (n); i++) p[i] = read(), cnt[(p[i] - i + n) % n]++;
    vec.clear();
    for (k = 0; k <= (n - 1); k++)
      if (cnt[k] >= n - (m << 1)) {
        for (i = 1; i <= (n); i++) to[((i + k) % n ? (i + k) % n : n)] = p[i];
        s = n;
        for (i = 1; i <= (n); i++)
          if (to[i]) {
            j = i;
            while (to[j]) {
              h = to[j];
              to[j] = 0;
              j = h;
            }
            s--;
          }
        if (s <= m) vec.push_back(k);
      }
    write(vec.size());
    if (!vec.empty() && !vec[0]) vec.erase(vec.begin()), vec.push_back(0);
    while (!vec.empty()) {
      putchar(' ');
      write((n - vec.back()) % n);
      vec.pop_back();
    }
    putchar('\n');
  }
  return 0;
}

```