#include<bits/stdc++.h>
using namespace std;
#define ran 111111
int n;
int a[ran],b[ran];

int N;
int A[ran],B[ran];

int n2;
map<int,int> Map;
int h(int x){
	if(Map.count(x))
		return Map[x];
	Map[x] = n2;
	n2++;
	return Map[x];
}

vector<pair<int,int> > e;
int ace[ran];
int ancestor(int x){return x-ace[x]?ace[x]=ancestor(ace[x]):x;}
int stat(){
	for(int i=0; i<n2; i++)
		ace[i] = i;
	for(vector<pair<int,int> >::iterator it = e.begin(); it!=e.end(); it++)
		ace[ancestor(it->first)]=ancestor(it->second);
	int res = 0;
	for(int i=0; i<n2; i++)
		if(ancestor(i)==i)
			res++;
	return res;
}

int main(){
	scanf("%d",&n);
	for(int i=0; i<n; i++){
		scanf("%d",&a[i]);
		a[n] ^= a[i];
	}
	for(int i=0; i<n; i++){
		scanf("%d",&b[i]);
		b[n] ^= b[i];
	}

	N = n+1;
	for(int i=0; i<N; i++){
		A[i] = a[i];
		B[i] = b[i];
	}
	sort(A, A+N);
	sort(B, B+N);
	bool flag = true;
	for(int i=0; i<N; i++)
		if(A[i] != B[i])
			flag = false;
	if(!flag){
		puts("-1");
		return 0;
	}

	for(int i=0; i<N; i++)
		if(a[i] != b[i]){
			int t1 = h(a[i]);
			int t2 = h(b[i]);
			e.push_back(make_pair(t1,t2));
		}

	int res = e.size();

	if(a[N-1] != b[N-1]){
		res += stat() - 2;
	}else{
		res += stat();
		if(Map.count(a[N-1]))
			res --;
	}
	printf("%d\n",res);
	return 0;
}

### 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 ran 111111
int n;
int a[ran],b[ran];

int N;
int A[ran],B[ran];

int n2;
map<int,int> Map;
int h(int x){
	if(Map.count(x))
		return Map[x];
	Map[x] = n2;
	n2++;
	return Map[x];
}

vector<pair<int,int> > e;
int ace[ran];
int ancestor(int x){return x-ace[x]?ace[x]=ancestor(ace[x]):x;}
int stat(){
	for(int i=0; i<n2; i++)
		ace[i] = i;
	for(vector<pair<int,int> >::iterator it = e.begin(); it!=e.end(); it++)
		ace[ancestor(it->first)]=ancestor(it->second);
	int res = 0;
	for(int i=0; i<n2; i++)
		if(ancestor(i)==i)
			res++;
	return res;
}

int main(){
	scanf("%d",&n);
	for(int i=0; i<n; i++){
		scanf("%d",&a[i]);
		a[n] ^= a[i];
	}
	for(int i=0; i<n; i++){
		scanf("%d",&b[i]);
		b[n] ^= b[i];
	}

	N = n+1;
	for(int i=0; i<N; i++){
		A[i] = a[i];
		B[i] = b[i];
	}
	sort(A, A+N);
	sort(B, B+N);
	bool flag = true;
	for(int i=0; i<N; i++)
		if(A[i] != B[i])
			flag = false;
	if(!flag){
		puts("-1");
		return 0;
	}

	for(int i=0; i<N; i++)
		if(a[i] != b[i]){
			int t1 = h(a[i]);
			int t2 = h(b[i]);
			e.push_back(make_pair(t1,t2));
		}

	int res = e.size();

	if(a[N-1] != b[N-1]){
		res += stat() - 2;
	}else{
		res += stat();
		if(Map.count(a[N-1]))
			res --;
	}
	printf("%d\n",res);
	return 0;
}

```