#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5;
int n,a[N],b[N];
map<int,int>p;
inline int bel(int i){
	return i==p[i]?i:p[i]=bel(p[i]);
}
int main(){
	cin>>n;
	multiset<int>c,d;
	for(int i=1;i<=n;++i)
		cin>>a[i],a[0]^=a[i],c.insert(a[i]);
	c.insert(a[0]);
	for(int i=1;i<=n;++i)
		cin>>b[i],b[0]^=b[i],d.insert(b[i]);
	d.insert(b[0]);
	if(c!=d)
		return!~puts("-1");
	for(int i=1;i<=n;++i)
		if(a[i]!=b[i])
			p[a[i]]=a[i];
	p[a[0]]=a[0];
	int s=-1;
	for(int i=1;i<=n;++i)
		if(a[i]!=b[i])
			p[bel(a[i])]=bel(b[i]),++s;
	for(auto e:p)
		if(e.first==e.second)++s;
	cout<<s<<endl;
}
### Prompt

Please provide a Cpp coded solution 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<bits/stdc++.h>
using namespace std;
const int N=1e5+5;
int n,a[N],b[N];
map<int,int>p;
inline int bel(int i){
	return i==p[i]?i:p[i]=bel(p[i]);
}
int main(){
	cin>>n;
	multiset<int>c,d;
	for(int i=1;i<=n;++i)
		cin>>a[i],a[0]^=a[i],c.insert(a[i]);
	c.insert(a[0]);
	for(int i=1;i<=n;++i)
		cin>>b[i],b[0]^=b[i],d.insert(b[i]);
	d.insert(b[0]);
	if(c!=d)
		return!~puts("-1");
	for(int i=1;i<=n;++i)
		if(a[i]!=b[i])
			p[a[i]]=a[i];
	p[a[0]]=a[0];
	int s=-1;
	for(int i=1;i<=n;++i)
		if(a[i]!=b[i])
			p[bel(a[i])]=bel(b[i]),++s;
	for(auto e:p)
		if(e.first==e.second)++s;
	cout<<s<<endl;
}
```