#include<bits/stdc++.h>
using namespace std;
#define fs first
#define sc second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define ALL(A) A.begin(),A.end()
#define RALL(A) A.rbegin(),A.rend()
typedef long long LL;
typedef pair<LL,LL> P;
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; }
template<typename T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
const LL mod=1000000007;
const LL LINF=1LL<<62;
const int INF=1<<30;
int dx[]={1,0,-1,0,1,-1,1,-1};
int dy[]={0,1,0,-1,1,-1,-1,1};

template< typename T >
struct BinaryIndexedTree{

    //0-indexed
    vector< T > bit;

    BinaryIndexedTree(int sz){
        bit.assign(sz, 0);
    }

    //[0,k)までの総和を返す
    T sum(int k){
        T ret = 0;
        for (--k; k >= 0; k = (k&(k+1))-1) ret += bit[k];
        return ret;
    }
    
    //[l,r)の総和を返す
    T sum(int l, int r){
        return sum(r) - sum(l);
    }

    //kにx加算する
    void add(int k, T x){
        for (; k < bit.size(); k |= k+1) bit[k] += x;
    }
};
 
int main(){
    int n;cin >> n;
    vector<int> a(n),b(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    for (int i = 0; i < n; i++) {
        cin >> b[i];
    }
    if(a[0] != b[0] || a[n-1] != b[n-1]){
        puts("-1");
        return 0;
    }
    vector<int> av(n-1), bv(n-1);
    map<int,vector<int> > ma,mb;
    for (int i = 0; i < n-1; i++) {
        av[i] = a[i] ^ a[i+1];
        ma[av[i]].pb(i);
        bv[i] = b[i] ^ b[i+1];
        mb[bv[i]].pb(i);
    }
    for(auto A:ma){
        if(A.sc.size() != mb[A.fs].size()){
            puts("-1");
            return 0;
        }
    }
    vector<int> map_av(n - 1);
    map<int,int> m_id;
    for (int i = 0; i < n-1; i++) {
        map_av[i] = mb[av[i]][m_id[av[i]]++];
    }
    BinaryIndexedTree<int> bit(n);
    LL ans = 0;
    for (int i = 0; i < n-1; i++) {
        ans += bit.sum(map_av[i], n);
        bit.add(map_av[i], 1);
    }
    cout << ans << endl;
    return 0;
}