#include<bits/stdc++.h>
using namespace std;
#define N 1111116

int n,k,a[N],b[N],A[N],B[N],f[N],ans;
map<int,int> mp;

int get(int x){
	return f[x]==x?x:f[x]=get(f[x]);
}

int read(){
	int x=0,f=1;char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-') f=-f;
	for (;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
	return x*f;
}

int main(){
	n=read();
	for (int i=1;i<=n;i++) a[n+1]^=(a[i]=read());
	for (int i=1;i<=n;i++) b[n+1]^=(b[i]=read());
	memcpy(A,a,sizeof(A));sort(A+1,A+n+2);
	memcpy(B,b,sizeof(B));sort(B+1,B+n+2);
	for (int i=1;i<=n+1;i++)
		if (A[i]!=B[i]) {printf("-1\n");return 0;};
	for (int i=1;i<=n;i++)
		if (a[i]!=b[i]){
			ans++;
			if (!mp[a[i]]) mp[a[i]]=++k;
			if (!mp[b[i]]) mp[b[i]]=++k;
		}
	if (!mp[a[n+1]]) mp[a[n+1]]=++k;
	if (!mp[b[n+1]]) mp[b[n+1]]=++k;
	for (int i=1;i<=k;i++) f[i]=i;
	for (int i=1;i<=n+1;i++)
		if (a[i]!=b[i]) f[get(mp[a[i]])]=get(mp[b[i]]);
	for (int i=1;i<=k;i++)
		if (f[i]==i) ans++;
	printf("%d\n",ans-1);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<bits/stdc++.h>
using namespace std;
#define N 1111116

int n,k,a[N],b[N],A[N],B[N],f[N],ans;
map<int,int> mp;

int get(int x){
	return f[x]==x?x:f[x]=get(f[x]);
}

int read(){
	int x=0,f=1;char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-') f=-f;
	for (;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
	return x*f;
}

int main(){
	n=read();
	for (int i=1;i<=n;i++) a[n+1]^=(a[i]=read());
	for (int i=1;i<=n;i++) b[n+1]^=(b[i]=read());
	memcpy(A,a,sizeof(A));sort(A+1,A+n+2);
	memcpy(B,b,sizeof(B));sort(B+1,B+n+2);
	for (int i=1;i<=n+1;i++)
		if (A[i]!=B[i]) {printf("-1\n");return 0;};
	for (int i=1;i<=n;i++)
		if (a[i]!=b[i]){
			ans++;
			if (!mp[a[i]]) mp[a[i]]=++k;
			if (!mp[b[i]]) mp[b[i]]=++k;
		}
	if (!mp[a[n+1]]) mp[a[n+1]]=++k;
	if (!mp[b[n+1]]) mp[b[n+1]]=++k;
	for (int i=1;i<=k;i++) f[i]=i;
	for (int i=1;i<=n+1;i++)
		if (a[i]!=b[i]) f[get(mp[a[i]])]=get(mp[b[i]]);
	for (int i=1;i<=k;i++)
		if (f[i]==i) ans++;
	printf("%d\n",ans-1);return 0;
}
```