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>
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("O3")
#pragma GCC target("avx,avx2,fma")
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, -1, 0, 1};
long long rnpow(long long x, long long y) {
  if (y == 0) return 1;
  if (y & 1) return rnpow(x, y - 1) * x % 1000000007LL;
  long long res = rnpow(x, y >> 1LL) % 1000000007LL;
  return res * res % 1000000007LL;
}
int gcd(int i, int j) { return i ? gcd(j % i, i) : j; }
long long gcd(long long i, long long j) { return i ? gcd(j % i, i) : j; }
long long fac(long long x) { return x ? fac(x - 1) * x : 1; }
int ar[300003], dp[300003];
std::pair<int, int> ap[300003];
int n, m;
std::vector<int> v;
int query(int x) {
  std::vector<std::pair<int, int> > w;
  for (int i = x + 1; i <= n; ++i) {
    if ((i + n - ar[i]) % n != x) w.emplace_back(ar[i], (int)w.size());
  }
  for (int i = 1; i <= x; ++i) {
    if ((i + n - ar[i]) % n != x) w.emplace_back(ar[i], (int)w.size());
  }
  std::sort((w).begin(), (w).end());
  std::vector<int> vs(w.size() + 1, 0);
  int res = 0;
  for (int i = 0; i < w.size(); ++i) {
    if (vs[i] || w[i].second == i) continue;
    int cs = 0;
    for (int j = i; !vs[j];) {
      vs[j] = 1;
      j = w[j].second;
      ++cs;
    }
    if (cs > 0) res += ~-cs;
  }
  return res;
}
int main() {
  int tc;
  for (scanf("%d", &tc); tc--;) {
    v.clear();
    scanf("%d%d", &n, &m);
    for (int i = 0; i <= n; ++i) dp[i] = 0;
    for (int i = 1; i <= n; ++i) {
      scanf("%d", ar + i);
      ++dp[(i + n - ar[i]) % n];
    }
    for (int i = 0; i < n; ++i) {
      if ((n - dp[i] + 1) / 2 <= m) {
        if (query(i) <= m) v.push_back(i);
      }
    }
    printf("%lu ", v.size());
    for (auto& it : v) printf("%d ", it);
    puts("");
  }
  return 0;
}
