// naive2
#include <cstdio>
#include <cassert>
#include <iostream>
#include <cstring>
using namespace std;
typedef long long ll;
#define N (1<<24)
#define M 1000
int m, n, mx, ax, my, ay, mod;
ll x[M], y[M], z[N], a[M], b[M], dp[N];

ll solve2(ll a){
	if(a>mod) return 0;
	if(dp[a]>=0) return dp[a];
	ll res = 0;
	for(ll i = a; i < mod; i+=a){
		res += z[i];
	}
	return dp[a] = res;
}

int main(){
	scanf("%d%d%d%d%d%d%d",&m,&n,&mx,&ax,&my,&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 = 0; i < m; i++) z[x[i]] += y[i];
	ll xx = x[m-1], yy = y[m-1];
	for(int i = m; i < n; i++){
		xx = (xx*mx+(ll)ax)&(mod-1);
		yy = (yy*my+(ll)ay)&(mod-1);
		z[xx] += yy;
	}
	
	for(int i = 0; i < m; i++) scanf("%d", a+i);
	for(int i = 0; i < m; i++) scanf("%d", b+i);

	memset(dp, -1, sizeof(dp));
	ll res = 0;
	for(int i = 0; i < m; i++){
		ll r = solve2(a[i])-solve2(a[i]*b[i]);
		printf("%lld\n", r);
		res^=r;
	}
	ll aa = a[m-1], bb = b[m-1];
	for(int i = m; i < n; i++){
		aa = ((aa*mx+(ll)ax+mod-1)&(mod-1))+1;
		bb = ((bb*my+(ll)ay+mod-1)&(mod-1))+1;
		res^=solve2(aa)-solve2(aa*bb);
	}
	printf("%lld\n", res);
	return 0;
}