「NOI2018」屠龙勇士

/*
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;
}