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]. 
#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 2;
int n, m, a[N], cnt[N], nxt[N];
bool vis[N];
int main() {
  int test;
  cin >> test;
  while (test--) {
    cin >> n >> m;
    for (int i = 1; i <= n; ++i) cnt[i] = 0;
    for (int i = 1; i <= n; ++i) {
      cin >> a[i];
      int t1 = i + n - a[i], t2 = i - a[i] + 1;
      int k1 = -1, k2 = -1;
      if (t1 < n) {
        k1 = t1;
      }
      if (t2 > 0) {
        k2 = t2 - 1;
      }
      if (k1 >= 0) ++cnt[k1];
      if (k2 >= 0) cnt[k2] += k1 != k2;
    }
    vector<int> ans;
    ans.clear();
    for (int i = 0; i < n; ++i) {
      if (n - cnt[i] > m * 2) continue;
      for (int j = 1; j <= n; ++j) {
        vis[j] = false;
        if (a[j] <= n - i) {
          nxt[j] = i + a[j];
        } else {
          nxt[j] = a[j] - (n - i);
        }
      }
      int cycles = 0;
      for (int j = 1; j <= n; ++j)
        if (nxt[j] != j && !vis[j]) {
          int u = j;
          ++cycles;
          vis[j] = true;
          while (nxt[u] != j) {
            u = nxt[u];
            vis[u] = true;
          }
        }
      if (n - cnt[i] - cycles <= m) {
        ans.push_back(i);
      }
    }
    cout << (int)ans.size() << " ";
    for (int i : ans) cout << i << " ";
    cout << endl;
  }
  return 0;
}
