#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<set>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
const int INF=0x3f3f3f3f;
typedef long long LL;
LL M[15],B[15];
int t;
LL n,m,p,num;
LL exgcd(LL a, LL b, LL &x, LL &y)
{
    if (!b)
    {
        x = 1;
        y = 0;
        return a;
    }
    LL gcd = exgcd(b, a % b, x, y);
    LL t = x;
    x = y;
    y = t - (a / b) * x;
    return gcd;
}
LL mult(LL a,LL b,LL p)
{
    LL ans=0;
    while(b)
    {
        if(b&1)
            ans=(ans+a)%p;
        b>>=1;
        a=(a+a)%p;
    }
    return ans;
}
LL inverse(LL num, LL mod)
{
    LL x, y;
    exgcd(num, mod, x, y);
    return (x % mod + mod) % mod;
}

LL C(LL a, LL b, LL mod)
{
    if (b > a)
        return 0;
    LL t1 = 1, t2 = 1;
    for (int i = 1; i <= b; ++i)
    {
        t2 *= i;
        t2 %= mod;
        t1 *= (a - i + 1);
        t1 %= mod;
    }
    return mult(t1,inverse(t2, mod),mod);
}
LL lucas(LL n, LL m, LL p)
{
    if (m == 0)
        return 1;
    return mult(C(n % p, m % p, p),lucas(n / p, m / p, p),p);
}
LL China(LL m[],LL b[],int k)//m:模数 b:余数
{
    LL n=1,xx,yy;
    LL ans=0;
    for(int i=0;i<k;i++)
        n*=M[i];
    for(int i=0;i<k;i++)
    {
        LL t=n/M[i];
        exgcd(t,M[i],xx,yy);
        LL x1=mult(xx,t,n);
        LL x2=mult(x1,b[i],n);
        ans=(ans+x2)%n;
    }
    return (ans%n+n)%n;
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld%lld%lld",&n,&m,&num);
        for(int i=0;i<num;i++)
        {
            scanf("%d",&p);
            M[i]=p;
            B[i]=lucas(n,m,p);
        }
        LL ans=China(M,B,num);
        printf("%lld\n",ans);
    }
    return 0;
}