#pragma GCC optimize("Ofast")
#include <cstdio>
#include <iostream>
#include <string>
#include <sstream>
#include <stack>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map>
#include <set>
#include <cstdlib>
#include <bitset>
#include <tuple>
#include <assert.h>
#include <deque>
#include <bitset>
#include <iomanip>
#include <limits>
#include <chrono>
#include <random>
#include <array>
#include <unordered_map>
#include <functional>
#include <complex>
#include <numeric>
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

const long long MAX = 5100000;
const long long INF = 1LL << 60;
//const long long mod = 1000000007LL;
//const long long mod = 998244353LL;

using namespace std;
typedef unsigned long long ull;
typedef long long ll;


ll A[150000];
ll B[150000];
int main()
{
	/*
	cin.tie(nullptr);
	ios::sync_with_stdio(false);
	*/
	ll N, M, K; scanf("%lld %lld %lld", &N, &M, &K);
	char c; cin >> c;
	for (ll i = 0; i < M; i++) scanf("%lld", &B[i]);
	for (ll i = 0; i < N; i++) scanf("%lld", &A[i]);
	sort(B, B + M);
	sort(A, A + N);
	//map<ll, ll> mp = prime_factor(K);
	ll res = 0;
	if (c == '+') {
		for (ll i = 0; i < M; i++) B[i] %= K;
		for (ll i = 0; i < N; i++) A[i] %= K;
		sort(B, B + M);
		sort(A, A + N);
		for (ll i = 0; i < N; i++) {
			if (A[i] == 0) {
				res += upper_bound(B, B + M, 0) - lower_bound(B, B + M, 0);
			}
			else {
				ll req = K - A[i];
				res += upper_bound(B, B + M, req) - lower_bound(B, B + M, req);
			}
		}
	}
	else {
		vector<ll> d;
		//d.reserve(1335);
		for (ll i = 1; i * i <= K; i++) {
			if (K % i == 0) {
				d.push_back(i);
				if (i != K / i) d.emplace_back(K / i);
			}
		}
		sort(d.begin(), d.end());
		vector<ll> cnt(d.size());
		for (ll i = 0; i < N; i++) for (ll j = 0; j < d.size(); j++) if (A[i] % d[j] == 0) cnt[j]++;
		for (ll i = 0; i < M; i++) {
			for (ll j = (ll)d.size() - 1; j >= 0; j--) {
				if (B[i] % d[j] == 0) {
					ll req = K / d[j];
					ll id = lower_bound(d.begin(), d.end(), req) - d.begin();
					res += cnt[id];
					break;
				}
			}
			//cout << res << endl;
		}
	}
	printf("%lld\n", res);
	return 0;
}