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;
#define ll long long  
#define pb push_back 
#define mk make_pair

ll m=1000000007; 
ll fact(ll n); 

int modInverse(int a, int m) 
{ 
    int m0 = m; 
    int y = 0, x = 1; 
  
    if (m == 1) 
      return 0; 
  
    while (a > 1) 
    { 
        int q = a / m; 
        int t = m; 
  
        m = a % m, a = t; 
        t = y; 
  
        y = x - q * y; 
        x = t; 
    } 
  
    if (x < 0) 
       x += m0; 
  
    return x; 
} 


ll fact(ll n) 
{ 
    ll res = 1; 
    for (ll i = 2; i <= n; i++) 
        res = (res%m * i%m)%m; 
    return res; 
} 
 vector<bool> mark;
  vector<bool> pmark;
  ll c;
 
void dfs(vector<vector<ll>> &v, ll node) {
    if (pmark[node]) {
      return;
    }
    if (mark[node]) {
      c++;
      return;
    }
    mark[node] = true;
 
    // Try all outgoing edges.
    for (ll i=0;i<v[node].size();i++) {
      dfs(v,v[node][i]);
    }
 
    pmark[node] = true;
    mark[node] = false;
  }


int main() {
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        ll t;
        cin>>t;
        while(t--){
            ll n,an=0,m;
            string s;
            cin>>n>>m;
            ll a[n];
             vector<ll>vv;
            ll b[n]={0};
            for(ll i=0;i<n;i++){
                cin>>a[i];
                ll x;
                if(a[i]-1<=i){
                    x=i-(a[i]-1);
                    b[x]++;
                }else{
                    x=n-(a[i]-1-i);
                    b[x]++;
                }
            }
            
            // for(ll i=0;i<n;i++)cout<<b[i]<<" ";
            
            for(ll i=0;i<n;i++){
                ll x=n-b[i];
                ll y=x/2,z=x-1;
                if(x%2==1)y++;
                if(m>=z){
                    vv.pb(i);
                    continue;
                }
                if(m>=y){
                    vector<vector<ll>>v(n+1);
            
                    for(ll j=0;j<n;j++){
                    ll x=(a[j]+i)%n;
                    
                    ll xx;
                if(a[j]-1<=j){
                    xx=n-(a[j]-1);
                   
                }else{
                    xx=n-(a[j]-1-j);
                  // b[x]++;
                }
                //cout<<x<<" ";
                    if(xx!=i)    
                    v[a[j]].pb(a[x]);
                    }
                    
                    mark = vector<bool>(v.size(), false);
                    pmark = vector<bool>(v.size(), false);
                    c=0;
                    for(ll ii=0;ii<v.size();ii++){
                        dfs(v,ii);
                    }
                    ll z=x-c;
                    if(m>=z){
                        vv.pb(i);
                    }
                }
            }
            cout<<vv.size()<<" ";
            for(ll i=0;i<vv.size();i++)cout<<vv[i]<<" ";
            cout<<endl;
        }
    return 0;
}