#include <bits/stdc++.h>
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
using namespace std;
const int maxn = 300000;
const int inf = 0x3f3f3f3f;
const int mod = 1000000007;
int a[maxn + 5];
int cnt[maxn + 5];
bool mark[maxn + 5];
bool test(int n, int m, int off) {
  auto cal = [&](int x) {
    int y = x + off;
    return y <= n ? y : y - n;
  };
  int ans = 0;
  for (auto i = (1); i <= (n); i++) mark[i] = 0;
  for (auto i = (1); i <= (n); i++)
    if (mark[a[i]] == 0) {
      int now = a[i], c = 0;
      while (mark[now] == 0) {
        mark[now] = 1;
        now = a[cal(now)];
        c++;
      }
      ans += c - 1;
    }
  return ans <= m;
}
int main() {
  int cas;
  scanf("%d", &cas);
  while (cas--) {
    int n, m;
    scanf("%d%d", &n, &m);
    for (auto i = (1); i <= (n); i++) scanf("%d", &a[i]);
    int fixed = n - m * 2;
    for (auto i = (0); i <= (n - 1); i++) cnt[i] = 0;
    for (auto i = (1); i <= (n); i++) cnt[(i - a[i] + n) % n]++;
    vector<int> ans;
    for (auto i = (0); i <= (n - 1); i++)
      if (cnt[i] >= fixed && test(n, m, i)) ans.push_back(i);
    int N = ans.size();
    printf("%d", N);
    for (auto x : ans) printf(" %d", x);
    puts("");
  }
  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>
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
using namespace std;
const int maxn = 300000;
const int inf = 0x3f3f3f3f;
const int mod = 1000000007;
int a[maxn + 5];
int cnt[maxn + 5];
bool mark[maxn + 5];
bool test(int n, int m, int off) {
  auto cal = [&](int x) {
    int y = x + off;
    return y <= n ? y : y - n;
  };
  int ans = 0;
  for (auto i = (1); i <= (n); i++) mark[i] = 0;
  for (auto i = (1); i <= (n); i++)
    if (mark[a[i]] == 0) {
      int now = a[i], c = 0;
      while (mark[now] == 0) {
        mark[now] = 1;
        now = a[cal(now)];
        c++;
      }
      ans += c - 1;
    }
  return ans <= m;
}
int main() {
  int cas;
  scanf("%d", &cas);
  while (cas--) {
    int n, m;
    scanf("%d%d", &n, &m);
    for (auto i = (1); i <= (n); i++) scanf("%d", &a[i]);
    int fixed = n - m * 2;
    for (auto i = (0); i <= (n - 1); i++) cnt[i] = 0;
    for (auto i = (1); i <= (n); i++) cnt[(i - a[i] + n) % n]++;
    vector<int> ans;
    for (auto i = (0); i <= (n - 1); i++)
      if (cnt[i] >= fixed && test(n, m, i)) ans.push_back(i);
    int N = ans.size();
    printf("%d", N);
    for (auto x : ans) printf(" %d", x);
    puts("");
  }
  return 0;
}

```