Problem:
Score : 1000 points

Problem Statement

There is a sequence of length N: a = (a_1, a_2, ..., a_N).
Here, each a_i is a non-negative integer.

Snuke can repeatedly perform the following operation:

Let the XOR of all the elements in a be x. Select an integer i (1 ≤ i ≤ N) and replace a_i with x.

Snuke's objective is to match a with another sequence b = (b_1, b_2, ..., b_N).
Here, each b_i is a non-negative integer.

Determine whether the objective is achievable, and find the minimum necessary number of operations if the answer is positive.

Constraints

2 ≤ N ≤ 10^5

a_i and b_i are integers.

0 ≤ a_i, b_i < 2^{30}

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N
b_1 b_2 ... b_N

Output

If the objective is achievable, print the minimum necessary number of operations.
Otherwise, print -1 instead.

Sample Input 1

3
0 1 2
3 1 0

Sample Output 1

2

At first, the XOR of all the elements of a is 3.
If we replace a_1 with 3, a becomes (3, 1, 2).

Now, the XOR of all the elements of a is 0.
If we replace a_3 with 0, a becomes (3, 1, 0), which matches b.

Sample Input 2

3
0 1 2
0 1 2

Sample Output 2

0

Sample Input 3

2
1 1
0 0

Sample Output 3

-1

Sample Input 4

4
0 1 2 3
1 0 3 2

Sample Output 4

5

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define MP make_pair
#define PB push_back
#define int long long
#define st first
#define nd second
#define rd third
#define FOR(i, a, b) for(int i =(a); i <=(b); ++i)
#define RE(i, n) FOR(i, 1, n)
#define FORD(i, a, b) for(int i = (a); i >= (b); --i)
#define REP(i, n) for(int i = 0;i <(n); ++i)
#define VAR(v, i) __typeof(i) v=(i)
#define FORE(i, c) for(VAR(i, (c).begin()); i != (c).end(); ++i)
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)(x).size())
using namespace std;
template<typename TH> void _dbg(const char* sdbg, TH h) { cerr<<sdbg<<"="<<h<<"\n"; }
template<typename TH, typename... TA> void _dbg(const char* sdbg, TH h, TA... t) {
  while(*sdbg != ',')cerr<<*sdbg++; cerr<<"="<<h<<","; _dbg(sdbg+1, t...);
}
#ifdef LOCAL
#define debug(...) _dbg(#__VA_ARGS__, __VA_ARGS__)
#define debugv(x) {{cerr <<#x <<" = "; FORE(itt, (x)) cerr <<*itt <<", "; cerr <<"\n"; }}
#else
#define debug(...) (__VA_ARGS__)
#define debugv(x)
#define cerr if(0)cout
#endif
#define next ____next
#define prev ____prev
#define left ____left
#define hash ____hash
typedef long long ll;
typedef long double LD;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<ll> VLL;
typedef vector<pair<int, int> > VPII;
typedef vector<pair<ll, ll> > VPLL;

template<class C> void mini(C&a4, C b4){a4=min(a4, b4); }
template<class C> void maxi(C&a4, C b4){a4=max(a4, b4); }
template<class T1, class T2>
ostream& operator<< (ostream &out, pair<T1, T2> pair) { return out << "(" << pair.first << ", " << pair.second << ")";}
template<class A, class B, class C> struct Triple { A first; B second; C third;
  bool operator<(const Triple& t) const { if (st != t.st) return st < t.st; if (nd != t.nd) return nd < t.nd; return rd < t.rd; } };
template<class T> void ResizeVec(T&, vector<int>) {}
template<class T> void ResizeVec(vector<T>& vec, vector<int> sz) {
  vec.resize(sz[0]); sz.erase(sz.begin()); if (sz.empty()) { return; }
  for (T& v : vec) { ResizeVec(v, sz); }
}
typedef Triple<int, int, int> TIII;
template<class A, class B, class C>
ostream& operator<< (ostream &out, Triple<A, B, C> t) { return out << "(" << t.st << ", " << t.nd << ", " << t.rd << ")"; }
template<class T> ostream& operator<<(ostream& out, vector<T> vec) { out<<"("; for (auto& v: vec) out<<v<<", "; return out<<")"; }

const int N = 1e5 + 5;
map<int, VI> slo;
map<int, bool> vis;
void Dfs(int a) {
  vis[a]++;
  for (auto nei : slo[a]) {
    if (vis[nei]) { continue; }
    Dfs(nei);
  }
}
int a[N], b[N];
int32_t main() {

  ios_base::sync_with_stdio(0);
  cout << fixed << setprecision(10);
  cerr << fixed << setprecision(10);
  cin.tie(0);
  //double beg_clock = 1.0 * clock() / CLOCKS_PER_SEC;
  
  int n;
  cin>>n;
  int ax = 0, bx = 0;
  vector<int> as, bs;
  RE (i, n) {
    cin>>a[i];
    as.PB(a[i]);
    ax ^= a[i];
  }
  a[n + 1] = ax;
  as.PB(ax);
  RE (i, n) {
    cin>>b[i];
    bs.PB(b[i]);
    bx ^= b[i];
  }
  b[n + 1] = bx;
  bs.PB(bx);
  sort(ALL(as)); sort(ALL(bs));
  if (as != bs) {
    cout<<"-1\n";
    return 0;
  }
  int mismatches = 0;
  RE (i, n + 1) {
    if (a[i] != b[i]) {
      mismatches++;
      slo[a[i]].PB(b[i]);
    }
  }
  debug(mismatches);
  if (mismatches == 0) {
    cout<<0<<endl;
    return 0;
  }
  if (!slo[a[n + 1]].empty()) {
    debug("on cyc");
    mismatches--;
  }
  if (a[n + 1] != b[n + 1]) {
    mismatches--;
  }
  int cycs = 0;
  RE (i, n + 1) {
    if (!slo[a[i]].empty() && vis[a[i]] == 0) {
      Dfs(a[i]);
      cycs++;
    }
  }
  debug(mismatches, cycs);
  cout<<mismatches + cycs<<endl;
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  return 0;
}