// naive solution
#include <cstdio>
#include <cassert>
using namespace std;
typedef long long ll;
#define N (1<<23)
int m, n, mx, ax, my, ay, mod, q;
ll x[N], y[N], z[N];
bool prime[N];

int main(){
	scanf("%d%d%d%d%d%d%d",&m,&n,&mx,&ax,&my,&ay,&mod);
	assert(1<=m&&m<=100);
	assert(m<=n&&n<=10000000);
	assert(1<=mod&&mod<=1<<23);
	assert(__builtin_popcount(mod)==1);
	assert(0<=mx&&mx<mod);
	assert(0<=ax&&ax<mod);
	assert(0<=my&&my<mod);
	assert(0<=ay&&ay<mod);
	for(int i = 0; i < m; i++) scanf("%d", x+i);
	for(int i = 0; i < m; i++) scanf("%d", y+i);
	for(int i = m; i < n; i++){
		x[i] = (x[i-1]*mx+ax)&(mod-1);
		y[i] = (y[i-1]*my+ay)&(mod-1);
	}
	for(int i = 0; i < n; i++){
		z[x[i]] += y[i];
		//printf("%lld %lld\n", x[i], y[i]);
	}
	for(int i = 2; i*i<N; i++){
		if(prime[i]) continue;
		for(int j = i*i; j<N; j+=i){
			prime[j] = true;
		}
	}
	scanf("%d", &q);
	assert(1<=q&&q<=100000);
	while(q--){
		int a, b;
		scanf("%d%d",&a,&b);
		assert(1<=a&&(ll)a<=1000000000/b&&a*b<=1000000000);
		ll res = 0;
		for(ll i = 1; i < mod; i++){
			if(i%a==0&&i%(a*b)!=0) res += z[i];
		}
		printf("%lld\n", res);
	}
}