#include <bits/stdc++.h>
using namespace std;
typedef signed long long ll;

#undef _P
#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
//-------------------------------------------------------


int N,M,K;
int A[202020],B[202020];
ll from[1024],to[1024];
ll S[1024],T[1024];
const ll mo=1000000007;

void solve() {
	int i,j,k,l,r,x,y; string s;
	
	cin>>N>>M>>K;
	FOR(i,N) {
		cin>>x;
		ZERO(to);
		from[0]++;
		FOR(j,1024) S[j^x]+=to[j^x]=from[j];
		swap(from,to);
	}
	ZERO(from);
	FOR(i,M) {
		cin>>x;
		ZERO(to);
		from[0]++;
		FOR(j,1024) T[j^x]+=to[j^x]=from[j];
		swap(from,to);
	}
	
	ll ret=0;
	FOR(i,1024) S[i]%=mo, T[i]%=mo;
	FOR(i,1024) (ret+=S[i]*T[i^K])%=mo;
	cout<<ret<<endl;
	
	
}


int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n'; FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	cout.tie(0); solve(); return 0;
}