#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
inline int read(){
	int x=0,f=1;char c=getchar();
	for(;!isdigit(c);c=getchar()) if(c==‘-‘) f=-1;
    for(;isdigit(c);c=getchar()) x=x*10+c-‘0‘;
    return x*f;
}
int gcd(int a,int b){
	if(b==0) return a;
    return gcd(b,a%b);
}
int ex_gcd(int a,int b,int &x,int &y){//扩展欧几里得
	if(b==0){
		x=1,y=0;
		return a;
	}
	int k=ex_gcd(b,a%b,x,y);
	int tmp=x;
	x=y;
	y=tmp-a/b*y;
	return k;
}
int T;
int N,M;
int w[11],r[11];
int China(int N){
	int M=w[1],R=r[1];
	int x,y;
	for(int i=2;i<=N;i++){
		int d=gcd(M,w[i]);
		int c=r[i]-R;
		if(c%d) {return -1;}
		ex_gcd(M/d,w[i]/d,x,y);
		x=(c/d*x)%(w[i]/d);
		R+=x*M;
		M=M/d*w[i];
		R%=M;
	}
	if(R<0) return R+M;
	else return R;
}
int main(){
	T=read();
	while(T--){
		N=read(),M=read();
		for(int i=1;i<=M;i++) w[i]=read();
		for(int i=1;i<=M;i++) r[i]=read();
		int ret=China(M);
		if(ret==-1||ret>N) {puts("0");continue;} //特判
		int lcm=1,ans=1;
		for(int i=1;i<=M;i++) lcm=lcm*w[i]/gcd(lcm,w[i]);//求所有数的最小公倍数,有个式子是a*b=gcd(a,b)*lcm(a*b);
		while(ret+lcm<N){
	        ans++;
			ret+=lcm; 
		}
		cout<<ans<<endl;
	}
}