#define _CRT_SECURE_NO_WARNINGS 1
#include<bits/stdc++.h>
#include<unordered_map>
#include<unordered_set>
using namespace std;
#define ll long long
#define double long double
#define lc u<<1
#define rc u<<1|1
#define X first
#define Y second
#define endl "\n"
//#define int long long
const int N = 2e5 + 50;
const int M = 2e5 + 5;
const ll maxm = 1e18 + 5;
const ll mod = 998244353;
int dx[] = { -1,0,1,0 };
int dy[] = { 0,1,0,-1 };
//random_shuffle(id + 1, id + n + 1);
ll ksm(ll a, ll b);
ll gcd(ll a, ll b);

template<class T>inline void read(T& x) {
	x = 0;
	char c = getchar();
	while (!isdigit(c))c = getchar();
	while (isdigit(c))x = x * 10 + (c & 15), c = getchar();
}

void write(ll x)
{
	if (x < 0)
		putchar('-'), x = -x;
	if (x > 9)
		write(x / 10);
	putchar(x % 10 + '0');
	return;
}

ll a[N], b[N];
ll c[N];

void solve()
{
	//这道赛时就出了，还有印象，我们发现应该优先利用较小的元素
	//如果能执行前置操作，那么我们就一直执行前置操作
	ll aa, bb; cin >> aa >> bb;
	for (int i = 1; i <= aa; i++)read(a[i]);
	for (int i = 1; i <= bb; i++)read(b[i]);
	if (bb > aa) {
		write(-1);
		puts(""); return;
	}
	sort(a + 1, a + 1 + aa);
	sort(b + 1, b + 1 + bb);
	ll bdis = 0;
	for (int i = 0; i < bb; i++) {
		if (b[bb - i] < a[aa - i]) {
			write(-1);
			puts(""); return;
		}
		bdis += b[bb - i] - a[aa - i];
	}
	ll sumdis = aa - bb;
	if (sumdis < bdis) {
		write(-1);
		puts(""); return;
	}

	vector<ll>ans;
	priority_queue<ll, vector<ll>, greater<ll> >bq;
	priority_queue<ll, vector<ll>, greater<ll> >fq;

	for (int i = 1; i <= aa; i++) {
		if (i <= aa - bb) {
			fq.push(a[i]);
		}
		else {
			bq.push(a[i]);
		}
	}
	while (sumdis > bdis) {
		if (fq.empty()) {
			write(-1);
			puts(""); return;
		}
		ll val = fq.top();
		fq.pop();
		ans.push_back(val); val++;
		if (val > bq.top()) {
			fq.push(bq.top());
			bq.pop();
			bq.push(val);
			bdis--;
			fq.pop();
		}
		else {
			fq.push(val);
			fq.pop();
		}
		sumdis--;
	}
	ll idx = 0;
	while (bq.size()) {
		c[++idx] = bq.top();
		bq.pop();
	}
	ll cnt = 0;
	for (int i = bb; i >= 1; i--) {
		if (c[i] > b[i]) {
			write(-1);
			puts(""); return;
		}
		for (int j = c[i]; j < b[i]; j++) {
			ans.push_back(j); cnt++;
		}
	}
	if (cnt != fq.size()) {
		write(-1);
		puts(""); return;
	}

	write(ans.size());
	puts("");

	for (int i = 0; i < ans.size(); i++) {
		write(ans[i]);
		putchar(' ');
	}
	puts("");
	return;
}


signed main()
{
	/*
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);*/
	int T = 1;
	//cin >> T;
	read(T);
	while (T--)
		solve();
	return 0;
}

/*PonyHex*/


ll ksm(ll a, ll b) {
	ll base = a;
	ll ans = 1;
	while (b) {
		if (b & 1)ans *= base % mod;
		ans %= mod;
		base *= base; base %= mod;
		b >>= 1;
	}
	return ans % mod;
}
ll gcd(ll a, ll b) {
	return b ? gcd(b, a % b) : a;
}