#include <bits/stdc++.h>
using namespace std;
const int max_n = 300005;
inline int read() {
  int x = 0;
  bool w = 0;
  char c = getchar();
  while (c < '0' || c > '9') w |= c == '-', c = getchar();
  while (c >= '0' && c <= '9')
    x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  return w ? -x : x;
}
inline void write(int x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 ^ 48);
}
struct graph {
  int ct, hd[max_n], nx[max_n * 2], to[max_n * 2];
  graph() { ct = 1; }
  inline void clear(int n) {
    ct = 1;
    for (register int i = 1; i <= n; ++i) hd[i] = 0;
  }
  inline void add(int u, int v) { nx[++ct] = hd[u], hd[u] = ct, to[ct] = v; }
} e;
int n, m, a[max_n], t[max_n];
bool vis[max_n];
inline void dfs(int x) {
  vis[x] = 1;
  for (register int i = e.hd[x]; i; i = e.nx[i]) {
    int v = e.to[i];
    if (vis[v]) continue;
    dfs(v);
  }
}
signed main() {
  int T = read();
  while (T--) {
    n = read(), m = read();
    for (register int i = 1; i <= n; ++i) {
      int x = read();
      a[x] = i;
      ++t[(i - x + n) % n];
    }
    int s = 0, ans[5];
    for (register int i = 0; i < n; ++i) {
      if (t[i] < n - 2 * m) continue;
      for (register int j = 1; j <= n; ++j) vis[j] = 0;
      e.clear(n);
      for (register int j = 1; j <= n; ++j) {
        int p = (i + j - 1) % n + 1;
        if (a[j] != p) e.add(p, a[j]);
      }
      int cnt = 0;
      for (register int j = 1; j <= n; ++j) {
        int p = (i + j - 1) % n + 1;
        if (a[j] != p && !vis[p]) {
          ++cnt;
          dfs(p);
        }
      }
      if (n - t[i] - cnt <= m) ans[++s] = i;
    }
    write(s), putchar(' ');
    for (register int i = 1; i <= s; ++i) write(ans[i]), putchar(' ');
    putchar('\n');
    for (register int i = 0; i <= n; ++i) t[i] = 0;
  }
  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>
using namespace std;
const int max_n = 300005;
inline int read() {
  int x = 0;
  bool w = 0;
  char c = getchar();
  while (c < '0' || c > '9') w |= c == '-', c = getchar();
  while (c >= '0' && c <= '9')
    x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  return w ? -x : x;
}
inline void write(int x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 ^ 48);
}
struct graph {
  int ct, hd[max_n], nx[max_n * 2], to[max_n * 2];
  graph() { ct = 1; }
  inline void clear(int n) {
    ct = 1;
    for (register int i = 1; i <= n; ++i) hd[i] = 0;
  }
  inline void add(int u, int v) { nx[++ct] = hd[u], hd[u] = ct, to[ct] = v; }
} e;
int n, m, a[max_n], t[max_n];
bool vis[max_n];
inline void dfs(int x) {
  vis[x] = 1;
  for (register int i = e.hd[x]; i; i = e.nx[i]) {
    int v = e.to[i];
    if (vis[v]) continue;
    dfs(v);
  }
}
signed main() {
  int T = read();
  while (T--) {
    n = read(), m = read();
    for (register int i = 1; i <= n; ++i) {
      int x = read();
      a[x] = i;
      ++t[(i - x + n) % n];
    }
    int s = 0, ans[5];
    for (register int i = 0; i < n; ++i) {
      if (t[i] < n - 2 * m) continue;
      for (register int j = 1; j <= n; ++j) vis[j] = 0;
      e.clear(n);
      for (register int j = 1; j <= n; ++j) {
        int p = (i + j - 1) % n + 1;
        if (a[j] != p) e.add(p, a[j]);
      }
      int cnt = 0;
      for (register int j = 1; j <= n; ++j) {
        int p = (i + j - 1) % n + 1;
        if (a[j] != p && !vis[p]) {
          ++cnt;
          dfs(p);
        }
      }
      if (n - t[i] - cnt <= m) ans[++s] = i;
    }
    write(s), putchar(' ');
    for (register int i = 1; i <= s; ++i) write(ans[i]), putchar(' ');
    putchar('\n');
    for (register int i = 0; i <= n; ++i) t[i] = 0;
  }
  return 0;
}

```