#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;

int u[108000];
int done[108000];
int dig[108000];
int cnt;

void ng(){
  cout << -1 << endl;
  exit(0);
}

int r(int x){
  if(x == u[x])return x;
  return u[x] = r(u[x]);
}

void unite(int x, int y){
  x = r(x);
  y = r(y);
  if(x != y)u[x] = y;
}

int n;

int main(){
  cin >> n;
  for(int i = 0;i < 108000;i++)u[i] = i;
  vector<int> a(n+1), b(n+1), c, d;
  a[n] = 0;
  for(int i = 0;i < n;i++){
    cin >> a[i];
    a[n] ^= a[i];
  }
  b[n] = 0;
  for(int i = 0;i < n;i++){
    cin >> b[i];
    b[n] ^= b[i];
  }
  
  c = a;
  d = b;
  sort(c.begin(), c.end());
  sort(d.begin(), d.end());
  if(c != d){
    ng();
  }

  c.erase(unique(c.begin(), c.end()), c.end());

  for(int i = 0;i <= n;i++){
    a[i] = lower_bound(c.begin(), c.end(), a[i]) - c.begin();
    b[i] = lower_bound(c.begin(), c.end(), b[i]) - c.begin();
  }
  

  for(int i = 0;i <= n;i++){
    if(a[i] == b[i])continue;
    unite(a[i], b[i]);
    dig[a[i]]++;
    cnt++;
  }
  for(int i = 0;i <= n;i++){
    if(a[i] == b[i])continue;
    if(done[r(a[i])])continue;
    done[r(a[i])] = true;
    cnt++;
  }
  if(a[n] != b[n])cnt -= 2;
  else if(dig[a[n]] > 0)cnt--;
  cout << cnt << endl;
  return 0;
}
### Prompt

Develop a solution in cpp to the problem described below:

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<iostream>
#include<algorithm>
#include<vector>
using namespace std;

int u[108000];
int done[108000];
int dig[108000];
int cnt;

void ng(){
  cout << -1 << endl;
  exit(0);
}

int r(int x){
  if(x == u[x])return x;
  return u[x] = r(u[x]);
}

void unite(int x, int y){
  x = r(x);
  y = r(y);
  if(x != y)u[x] = y;
}

int n;

int main(){
  cin >> n;
  for(int i = 0;i < 108000;i++)u[i] = i;
  vector<int> a(n+1), b(n+1), c, d;
  a[n] = 0;
  for(int i = 0;i < n;i++){
    cin >> a[i];
    a[n] ^= a[i];
  }
  b[n] = 0;
  for(int i = 0;i < n;i++){
    cin >> b[i];
    b[n] ^= b[i];
  }
  
  c = a;
  d = b;
  sort(c.begin(), c.end());
  sort(d.begin(), d.end());
  if(c != d){
    ng();
  }

  c.erase(unique(c.begin(), c.end()), c.end());

  for(int i = 0;i <= n;i++){
    a[i] = lower_bound(c.begin(), c.end(), a[i]) - c.begin();
    b[i] = lower_bound(c.begin(), c.end(), b[i]) - c.begin();
  }
  

  for(int i = 0;i <= n;i++){
    if(a[i] == b[i])continue;
    unite(a[i], b[i]);
    dig[a[i]]++;
    cnt++;
  }
  for(int i = 0;i <= n;i++){
    if(a[i] == b[i])continue;
    if(done[r(a[i])])continue;
    done[r(a[i])] = true;
    cnt++;
  }
  if(a[n] != b[n])cnt -= 2;
  else if(dig[a[n]] > 0)cnt--;
  cout << cnt << endl;
  return 0;
}
```