#include <bits/stdc++.h>
using namespace std;

#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define FORD(i,b,a) for (int i = (int)(b) - 1; i >= a; --i)
#define REP(i,N) FOR(i,0,N)
#define st first
#define nd second
#define pb push_back

typedef pair<int, int> PII;
typedef long long LL;

int A[111111], B[111111];

map<int, vector<int>> adj;
set<int> visited;

void go(int v) {
  visited.insert(v);
  for (auto u: adj[v]) if (visited.find(u) == visited.end()) go(u);
}

int main() {
  int N;
  scanf("%d", &N);
  int s = 0;
  REP(i,N) {
    scanf("%d", &A[i]);
    s ^= A[i];
  }
  A[N] = s;

  s = 0;
  REP(i,N) {
    scanf("%d", &B[i]);
    s ^= B[i];
  }
  B[N] = s;

  multiset<int> S;
  REP(i,N+1) S.insert(A[i]);
  REP(i,N+1) {
    if (S.find(B[i]) == S.end()) {
      printf("-1\n");
      return 0;
    }
    S.erase(S.find(B[i]));
  }

  int result = 0;
  REP(i,N) {
    if (A[i] != B[i]) {
      adj[B[i]].pb(A[i]);
      ++result;
    }
  }
  if (result == 0) {
    printf("0\n");
    return 0;
  }
  ++result;
  adj[B[N]].pb(A[N]);

  for(auto& p: adj) {
    if (visited.find(p.st) == visited.end()) {
      ++result;
      go(p.st);
    }
  }
  printf("%d\n", result - 2);

}

### Prompt

Generate a CPP solution to the following problem:

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.

Examples

Input

3
0 1 2
3 1 0


Output

2


Input

3
0 1 2
0 1 2


Output

0


Input

2
1 1
0 0


Output

-1


Input

4
0 1 2 3
1 0 3 2


Output

5

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define FORD(i,b,a) for (int i = (int)(b) - 1; i >= a; --i)
#define REP(i,N) FOR(i,0,N)
#define st first
#define nd second
#define pb push_back

typedef pair<int, int> PII;
typedef long long LL;

int A[111111], B[111111];

map<int, vector<int>> adj;
set<int> visited;

void go(int v) {
  visited.insert(v);
  for (auto u: adj[v]) if (visited.find(u) == visited.end()) go(u);
}

int main() {
  int N;
  scanf("%d", &N);
  int s = 0;
  REP(i,N) {
    scanf("%d", &A[i]);
    s ^= A[i];
  }
  A[N] = s;

  s = 0;
  REP(i,N) {
    scanf("%d", &B[i]);
    s ^= B[i];
  }
  B[N] = s;

  multiset<int> S;
  REP(i,N+1) S.insert(A[i]);
  REP(i,N+1) {
    if (S.find(B[i]) == S.end()) {
      printf("-1\n");
      return 0;
    }
    S.erase(S.find(B[i]));
  }

  int result = 0;
  REP(i,N) {
    if (A[i] != B[i]) {
      adj[B[i]].pb(A[i]);
      ++result;
    }
  }
  if (result == 0) {
    printf("0\n");
    return 0;
  }
  ++result;
  adj[B[N]].pb(A[N]);

  for(auto& p: adj) {
    if (visited.find(p.st) == visited.end()) {
      ++result;
      go(p.st);
    }
  }
  printf("%d\n", result - 2);

}

```