#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;
inline int read()
{
	int k=0,f=1;char c=getchar();
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) k=k*10+c-'0';return f*k;
}
const int N=100055;
map<int,int> mp;
int n,m,tot,a[N],b[N],c[N],d[N],as,fa[N<<1];
int find(int x)
{
	return fa[x]==x?x:fa[x]=find(fa[x]);
}
int main()
{
	n=read();
	for(int i=1;i<=n;i++)
		a[i]=read(),a[n+1]^=a[i];
	for(int i=1;i<=n;i++)
		b[i]=read(),b[n+1]^=b[i];
	n++;
	for(int i=1;i<=n;i++)
		c[i]=a[i],d[i]=b[i];
	sort(c+1,c+1+n);sort(d+1,d+1+n);
	for(int i=1;i<=n;i++)
		if(c[i]!=d[i])
			{puts("-1");return 0;}
	for(int i=1;i<=n;i++)
		if(a[i]!=b[i]||i==n)
		{
			if(i<n) as++;
			if(!mp[a[i]]) mp[a[i]]=++tot;
			if(!mp[b[i]]) mp[b[i]]=++tot;
		}
	if(!as) {puts("0");return 0;}
	for(int i=1;i<=tot;i++)
		fa[i]=i;
	for(int i=1;i<=n;i++)
		if(a[i]!=b[i])
			fa[find(mp[a[i]])]=find(mp[b[i]]);
	for(int i=1;i<=tot;i++)
		if(fa[i]==i)
			as++;
	printf("%d\n",as-1);
	return 0;
}
### Prompt

Please formulate 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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;
inline int read()
{
	int k=0,f=1;char c=getchar();
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) k=k*10+c-'0';return f*k;
}
const int N=100055;
map<int,int> mp;
int n,m,tot,a[N],b[N],c[N],d[N],as,fa[N<<1];
int find(int x)
{
	return fa[x]==x?x:fa[x]=find(fa[x]);
}
int main()
{
	n=read();
	for(int i=1;i<=n;i++)
		a[i]=read(),a[n+1]^=a[i];
	for(int i=1;i<=n;i++)
		b[i]=read(),b[n+1]^=b[i];
	n++;
	for(int i=1;i<=n;i++)
		c[i]=a[i],d[i]=b[i];
	sort(c+1,c+1+n);sort(d+1,d+1+n);
	for(int i=1;i<=n;i++)
		if(c[i]!=d[i])
			{puts("-1");return 0;}
	for(int i=1;i<=n;i++)
		if(a[i]!=b[i]||i==n)
		{
			if(i<n) as++;
			if(!mp[a[i]]) mp[a[i]]=++tot;
			if(!mp[b[i]]) mp[b[i]]=++tot;
		}
	if(!as) {puts("0");return 0;}
	for(int i=1;i<=tot;i++)
		fa[i]=i;
	for(int i=1;i<=n;i++)
		if(a[i]!=b[i])
			fa[find(mp[a[i]])]=find(mp[b[i]]);
	for(int i=1;i<=tot;i++)
		if(fa[i]==i)
			as++;
	printf("%d\n",as-1);
	return 0;
}
```