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 fast            ios_base::sync_with_stdio(false); cin.tie(NULL);
#define lli             long long int
#define ld              long double
#define pll             pair<lli,lli>
#define ve              vector<lli>
#define vep             vector<pll>
#define pb              push_back
#define mp              make_pair
#define mt              make_tuple
#define all(x)          (x).begin(),(x).end()
#define fr(i,a,n)       for(lli i=a;i<(n);i++)
#define frx(i,a,n)      for(lli i=a;i>(n);i--)
#define lb              lower_bound
#define ub              upper_bound
#define F               first
#define S               second
#define endl            '\n'
#define op(x)           cout<<x<<endl
#define op2(x,y)        cout<<x<<" "<<y<<endl
#define parr(arr,a,n)   fr(u,a,n) cout<<arr[u]<<" "; cout<<endl
#define yes             cout<<"YES"<<endl
#define no              cout<<"NO"<<endl
#define mem(a,x)        memset(a,x,sizeof(a))
#define sz(a) 		    ((lli) a.size())
#define db(x)           cout << "[ " << x << " ]" << endl
#define db2(x,y)        cout<<"[ " <<x<<" "<<y<< " ]" <<endl

void solve(){
    lli n,m;
    cin>>n>>m;
    lli a[n+5],b[n+5]={0};
    fr(i,1,n+1)cin>>a[i];
    fr(i,1,n+1){
        if(i>=a[i]) b[i-a[i]]++;
        else{
            b[n-a[i]+i]++;
        }
    }
    ve v;
    fr(i,0,n){
        lli temp=n-b[i];
        if((temp+1)/2<=m)v.pb(i);
    }
    cout<<sz(v)<<" ";
    parr(v,0,sz(v));
}

int main(){
    fast;
    lli test=1;
    cin>>test;
    while(test--){
        solve();
    }
}
