#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
template <typename T>
T gcd(T a,T b){
  if(a%b==0)return b;
  else return gcd(b,a%b);
}
template <typename T>
T lcm(T a,T b){
  return a/gcd(a,b)*b;
}
int main(){
    long long n,m,k;cin>>n>>m>>k;
    char z;cin>>z;
    vector<long long> A(n);
    vector<long long> B(m);
    for(int i = 0; m > i; i++)cin>>B[i];
    for(int i = 0; n > i; i++)cin>>A[i];
    //A[i]と互いに素なB[j]の個数が答え..........O(N^2)!!!!!!解散！！！！！笑
    if(z == '*'){
      for(int i = 0; m > i; i++){
        B[i] = gcd(B[i],k);
      }
      for(int i = 0; n > i; i++){
        A[i] = gcd(A[i],k);
      }
      vector<long long> nya;
      for(long long i = 1; k >= i*i; i++){
        if(!(k%i)){
          nya.push_back(i);
          if(i*i != k)nya.push_back(k/i);
        }
      }
      sort(B.begin(),B.end());
      long long ans = 0LL;
      for(int i = 0; n > i; i++){
        for(int j = 0; nya.size() > j; j++){
          if((nya[j]*A[i])%k)continue;
          ans += (upper_bound(B.begin(),B.end(),nya[j])-lower_bound(B.begin(),B.end(),nya[j]));
        }
      }
      cout << ans << endl;
    }else{
      long long ans = 0;
      for(int i = 0; m > i; i++){
        B[i] = B[i]%k;
      }
      sort(B.begin(),B.end());
      for(int i = 0; n > i; i++){
        A[i] = A[i]%k;
        int want = (k-A[i])%k;
        ans += (upper_bound(B.begin(),B.end(),want)-lower_bound(B.begin(),B.end(),want));
      }
      cout << ans << endl;
    }
}