/*
首先杀每条龙用到的刀是能够确定的， 然后我们便得到了许多形如 ai - x * atki | pi的方程
而且限制了x的最小值
那么exgcd解出来就好了
之后就是扩展crt合并了
因为全T调了一个小时 结果是没加文件？？
*/
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<set>
#include<iostream>
#define ll long long
#define M 100010
#define mmp make_pair
using namespace std;
ll read()
{
ll nm = 0, f = 1;
char c = getchar();
for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
return nm * f;
}
ll n, q, a[M], m[M], p[M], g[M], atk[M], tp, maxx;
multiset<ll> st;
ll mul(ll a, ll b, ll mod)
{
b = (b % mod + mod) % mod;
ll ans = 0, tmp = a;
for(; b; b >>= 1, tmp = (tmp + tmp) % mod) if(b & 1) ans = (ans + tmp) % mod;
return ans;
}
ll gcd(ll a, ll b)
{
return !b ? a : gcd(b, a % b);
}
ll exgcd(ll a, ll b, ll &x, ll &y)
{
if(!b)
{
x = 1, y = 0;
return a;
}
else
{
ll d = exgcd(b, a % b, x, y);
ll tmp = x;
x = y, y = tmp - a / b * y;
return d;
}
}
ll inv(ll a, ll p)
{
ll x, y;
ll d = exgcd(a, p, x, y);
if(d != 1) return -1;
return (x % p + p) % p;
}
void init()
{
st.clear();
tp = maxx = 0;
n = read(), q = read();
for(int i = 1; i <= n; i++) a[i] = read();
for(int i = 1; i <= n; i++) p[i] = read();
for(int i = 1; i <= n; i++) g[i] = read();
for(int i = 1; i <= q; i++) st.insert(read());
for(int i = 1; i <= n; i++)
{
multiset<ll>::iterator it = st.upper_bound(a[i]);
if(it != st.begin()) it--;
atk[i] = *it;
st.erase(it);
st.insert(g[i]);
}
}
ll excrt()
{
ll a1 = a[1], m1 = m[1], a2, m2;
if(tp == 0)
{
a1 = 0;
}
else
{
for(int i = 2; i <= tp; i++)
{
a2 = a[i], m2 = m[i];
ll d = gcd(m1, m2);
ll c = a2 - a1;
if(c % d) return -1;
ll k = inv(m1 / d, m2 / d);
m2 = m1 / d * m2;
a1 = mul(mul(m1 / d, c, m2), k, m2) + a1;
a1 %= m2;
m1 = m2;
}
}
return max(a1, maxx);
}
void cz(int i)
{
// a[i] - x * atk[i] + k * pi = 0
// a[i] = x * atk[i] - k * p[i]
// x * atk[i] = a[i] mod p[i]
//先处理gcd， 然后处理逆元
if(p[i] == 1)
{
maxx = max(maxx, (a[i] + atk[i] - 1) / atk[i]);
}
else
{
tp++;
ll gcdd = gcd(atk[i], p[i]);
if(a[i] % gcdd)
{
a[tp] = -1;
}
else
{
atk[i] /= gcdd, p[i] /= gcdd;
a[i] /= gcdd;
a[tp] = mul(a[i], inv(atk[i], p[i]), p[i]);
m[tp] = p[i];
}
}
}
void work()
{
for(int i = 1; i <= n; i++)
{
cz(i);
if(a[tp] == -1)
{
puts("-1");
return;
}
}
cout << excrt() << "\n";
}
int main()
{
freopen("dragon.in", "r", stdin);
freopen("dragon.out", "w", stdout);
//freopen("dragon1.in", "r", stdin);
int t = read();
while(t--)
{
init();
work();
}
return 0;
}