#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second 
#define endl '\n'
const LL maxn = 4e05+7;
const LL N=1e05+10;
const LL mod=1e09+7;
typedef pair<int,int>pl;
priority_queue<LL , vector<LL>, greater<LL> >t;
priority_queue<LL> q;
LL gcd(LL a, LL b){
	return b > 0 ? gcd(b , a % b) : a;
}

LL lcm(LL a , LL b){
	return a / gcd(a , b) * b;
}
void solve() 
{
	int n , m;
	cin>>n>>m;
	int a[n] , b[m] , c[n];
	int op = n - m;
	vector<int>res_a , res_b;
	for(int i = 0 ; i < n ; i++)
		cin>>a[i];
	for(int i = 0 ; i < m ; i ++)
		cin>>b[i];
	for(int i = 0 ; i < n ; i ++){
		c[i] = a[i];
	}
	LL cha = 0;//用最后m个数作为最终答案需要的操作数
	for(int i = 0 ; i < m ; i ++){
		if(a[i + (n - m)] > b[i]){
			cout << -1 <<endl;
			return;
		}
		else{
			cha += b[i] - a[i + (n - m)];
		}
	}
	//保留后 m 个数
	//最多前面 n - m 个数能有多少个无效操作
	LL max_none = 0;//一个加数最多能到多少
	LL st = a[n - m];
	vector<int>out;
	int i = 0 , r = n - 1;
	a[n] = 2e9;
	unordered_map<int,int>mp;
	for(i ; i < n - m ; i ++){
		if(a[i] < a[i + 1]){
			max_none ++;
			out.pb(a[i]);
		}
		else if(a[i] == a[i + 1]){//处理相等
			if(a[i] == b[0]){
				break;
			}
			else{
				while(a[r] <= a[i]) r++;
				while(a[r] > a[i]) r--;
				out.pb(a[r]);		
				mp[i] = r;
				if(r < n - m)
					max_none++;			
				a[r] ++;
			}
		}
	}
	if(cha > op || cha < op - max_none){
		cout<<-1<<endl;
	}
	else{
		cout<<op<<endl;
		int need = op - cha;
		for(int i = 0 ; i < need && i < out.size(); i ++){
			cout<<out[i]<<" ";
			if(mp[i] >= n - m){
				c[mp[i]]++;
				need += 1;
			}
		}
		for(int i = n - 1 ;i >= n - m ; i --){
			while(c[i] < b[i - n + m]){
				cout<<c[i]<<" ";
				c[i]++;
			}
		}
		cout<<endl;
	}
}            
int main() 
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cout.precision(10);
    int t=1;
	cin>>t;
    while(t--)
    {
    	solve();
    }
    return 0;
}