#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
using vii = vector<pii>;
using vll = vector<pll>;
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
vector<int> sieve;
void calc_sieve(int n) {
  sieve.reserve(n + 1);
  for (int i = 0, nnnn = (n + 1); i < nnnn; ++i) sieve[i] = i;
  for (int i = 2, nnnn = (n + 1); i < nnnn; ++i) {
    if (sieve[i] == i) {
      for (ll j = i * i; j <= n; j += i) {
        if (sieve[j] == j) sieve[j] = i;
      }
    }
  }
}
ll norm(ll a, ll P) {
  while (a >= P) a -= P;
  while (a < 0) a += P;
  return a;
}
ll mul_mod(ll a, ll b, ll P) { return a * b % P; }
ll pow_mod(ll a, ll b, ll P) {
  ll curr = a, ans = 1;
  while (b != 0) {
    if (b & 1) {
      ans *= curr;
      ans %= P;
    }
    curr *= curr;
    curr %= P;
    b >>= 1;
  }
  return ans;
}
ll inv_mod(ll a, ll P) { return pow_mod(a, P - 2ll, P); }
ll div_mod(ll a, ll b, ll P) { return a * inv_mod(b, P) % P; }
int msb(ll a) {
  for (int i = 62; i >= 0; i--) {
    if (a & ((ll)1 << i)) return i + 1;
  }
  return 0;
}
vector<ll> fact;
void calc_fact(int n, ll P) {
  if (fact.size() < 1) fact.push_back(1);
  for (int i = fact.size(), nnnn = (n + 1); i < nnnn; ++i)
    fact.push_back(fact[i - 1] * i % P);
}
vector<ll> fact_inv;
void calc_fact_inv(int n, ll P) {
  if (fact_inv.size() < 1) fact_inv.push_back(1);
  for (int i = fact_inv.size(), nnnn = (n + 1); i < nnnn; ++i)
    fact_inv.push_back(fact_inv[i - 1] * inv_mod(i, P) % P);
}
ll nCr(int n, int r, ll P) {
  if (r > n) return 0;
  calc_fact(n, P);
  calc_fact_inv(n, P);
  return (fact[n] * fact_inv[r] % P) * fact_inv[n - r] % P;
}
ll nCr_inv(int n, int r, ll P) {
  if (r > n) return 0;
  calc_fact(n, P);
  calc_fact_inv(n, P);
  return (fact_inv[n] * fact[r] % P) * fact[n - r] % P;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long double eps = 1e-9;
const long long mod = 1000000007;
const int MAXN = 100000;
void solve() {
  int n, m;
  cin >> n >> m;
  int p[n], diff[n];
  for (int i = 0, nnnn = (n); i < nnnn; ++i) {
    cin >> p[i], diff[i] = p[i] - i - 1;
    if (diff[i] < 0) diff[i] += n;
  }
  int cnt[n];
  memset(cnt, 0, sizeof cnt);
  for (int i = 0, nnnn = (n); i < nnnn; ++i) cnt[diff[i]]++;
  int ans = 0;
  for (int i = 0, nnnn = (n); i < nnnn; ++i) {
    if (n - cnt[i] <= m * 2) {
      int diff1[n], count = 0;
      for (int j = 0, nnnn = (n); j < nnnn; ++j) diff1[j] = diff[j];
      int j = 0;
      while (j < n) {
        if (diff1[j] % n != i) {
          int pos = (j + n + diff1[j] - i) % n;
          diff1[j] = (diff1[pos] + n - j + pos) % n;
          diff1[pos] = i;
          count++;
        }
        if (diff1[j] == i) j++;
      }
      if (count <= m) ans++;
    }
  }
  cout << ans << ' ';
  if (n - cnt[0] <= m * 2) {
    int diff1[n], count = 0;
    for (int j = 0, nnnn = (n); j < nnnn; ++j) diff1[j] = diff[j];
    int j = 0;
    while (j < n) {
      if (diff1[j] % n != 0) {
        int pos = (j + n + diff1[j] - 0) % n;
        diff1[j] = (diff1[pos] + n - j + pos) % n;
        diff1[pos] = 0;
        count++;
      }
      if (diff1[j] == 0) j++;
    }
    if (count <= m) cout << 0 << ' ';
  }
  for (int i = n - 1; i >= 1; i--) {
    if (n - cnt[i] <= m * 2) {
      int diff1[n], count = 0;
      for (int j = 0, nnnn = (n); j < nnnn; ++j) diff1[j] = diff[j];
      int j = 0;
      while (j < n) {
        if (diff1[j] % n != i) {
          int pos = (j + n + diff1[j] - i) % n;
          diff1[j] = (diff1[pos] + n - j + pos) % n;
          diff1[pos] = i;
          count++;
        }
        if (diff1[j] == i) j++;
      }
      if (count <= m) cout << n - i << ' ';
    }
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << setprecision(12) << fixed;
  int t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Please create a solution in cpp to 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;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
using vii = vector<pii>;
using vll = vector<pll>;
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
vector<int> sieve;
void calc_sieve(int n) {
  sieve.reserve(n + 1);
  for (int i = 0, nnnn = (n + 1); i < nnnn; ++i) sieve[i] = i;
  for (int i = 2, nnnn = (n + 1); i < nnnn; ++i) {
    if (sieve[i] == i) {
      for (ll j = i * i; j <= n; j += i) {
        if (sieve[j] == j) sieve[j] = i;
      }
    }
  }
}
ll norm(ll a, ll P) {
  while (a >= P) a -= P;
  while (a < 0) a += P;
  return a;
}
ll mul_mod(ll a, ll b, ll P) { return a * b % P; }
ll pow_mod(ll a, ll b, ll P) {
  ll curr = a, ans = 1;
  while (b != 0) {
    if (b & 1) {
      ans *= curr;
      ans %= P;
    }
    curr *= curr;
    curr %= P;
    b >>= 1;
  }
  return ans;
}
ll inv_mod(ll a, ll P) { return pow_mod(a, P - 2ll, P); }
ll div_mod(ll a, ll b, ll P) { return a * inv_mod(b, P) % P; }
int msb(ll a) {
  for (int i = 62; i >= 0; i--) {
    if (a & ((ll)1 << i)) return i + 1;
  }
  return 0;
}
vector<ll> fact;
void calc_fact(int n, ll P) {
  if (fact.size() < 1) fact.push_back(1);
  for (int i = fact.size(), nnnn = (n + 1); i < nnnn; ++i)
    fact.push_back(fact[i - 1] * i % P);
}
vector<ll> fact_inv;
void calc_fact_inv(int n, ll P) {
  if (fact_inv.size() < 1) fact_inv.push_back(1);
  for (int i = fact_inv.size(), nnnn = (n + 1); i < nnnn; ++i)
    fact_inv.push_back(fact_inv[i - 1] * inv_mod(i, P) % P);
}
ll nCr(int n, int r, ll P) {
  if (r > n) return 0;
  calc_fact(n, P);
  calc_fact_inv(n, P);
  return (fact[n] * fact_inv[r] % P) * fact_inv[n - r] % P;
}
ll nCr_inv(int n, int r, ll P) {
  if (r > n) return 0;
  calc_fact(n, P);
  calc_fact_inv(n, P);
  return (fact_inv[n] * fact[r] % P) * fact[n - r] % P;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long double eps = 1e-9;
const long long mod = 1000000007;
const int MAXN = 100000;
void solve() {
  int n, m;
  cin >> n >> m;
  int p[n], diff[n];
  for (int i = 0, nnnn = (n); i < nnnn; ++i) {
    cin >> p[i], diff[i] = p[i] - i - 1;
    if (diff[i] < 0) diff[i] += n;
  }
  int cnt[n];
  memset(cnt, 0, sizeof cnt);
  for (int i = 0, nnnn = (n); i < nnnn; ++i) cnt[diff[i]]++;
  int ans = 0;
  for (int i = 0, nnnn = (n); i < nnnn; ++i) {
    if (n - cnt[i] <= m * 2) {
      int diff1[n], count = 0;
      for (int j = 0, nnnn = (n); j < nnnn; ++j) diff1[j] = diff[j];
      int j = 0;
      while (j < n) {
        if (diff1[j] % n != i) {
          int pos = (j + n + diff1[j] - i) % n;
          diff1[j] = (diff1[pos] + n - j + pos) % n;
          diff1[pos] = i;
          count++;
        }
        if (diff1[j] == i) j++;
      }
      if (count <= m) ans++;
    }
  }
  cout << ans << ' ';
  if (n - cnt[0] <= m * 2) {
    int diff1[n], count = 0;
    for (int j = 0, nnnn = (n); j < nnnn; ++j) diff1[j] = diff[j];
    int j = 0;
    while (j < n) {
      if (diff1[j] % n != 0) {
        int pos = (j + n + diff1[j] - 0) % n;
        diff1[j] = (diff1[pos] + n - j + pos) % n;
        diff1[pos] = 0;
        count++;
      }
      if (diff1[j] == 0) j++;
    }
    if (count <= m) cout << 0 << ' ';
  }
  for (int i = n - 1; i >= 1; i--) {
    if (n - cnt[i] <= m * 2) {
      int diff1[n], count = 0;
      for (int j = 0, nnnn = (n); j < nnnn; ++j) diff1[j] = diff[j];
      int j = 0;
      while (j < n) {
        if (diff1[j] % n != i) {
          int pos = (j + n + diff1[j] - i) % n;
          diff1[j] = (diff1[pos] + n - j + pos) % n;
          diff1[pos] = i;
          count++;
        }
        if (diff1[j] == i) j++;
      }
      if (count <= m) cout << n - i << ' ';
    }
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << setprecision(12) << fixed;
  int t;
  cin >> t;
  while (t--) solve();
}

```