#include <iostream> #include <vector> #include <algorithm> #include <map> 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; long long ans = 0; 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(NM)!!!!!!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> C; for(long long i = 1; k >= i*i; i++){ if(!(k%i)){ C.push_back(i); if(i*i != k)C.push_back(k/i); } } map<int,int> nya; for(int i = 0; m > i; i++){ for(int j = 0; C.size() > j; j++){ if(!(B[i]%C[j])){ nya[C[j]]++; } } } for(int i = 0; n > i; i++){ ans += nya[k/A[i]]; } cout << ans << endl; }else{//O(n log m) 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; } }