#include<cstdio>
#include<map>
#include<algorithm>
#define MAXN 100006
using namespace std;
map<int,int> ma;
int a[MAXN],b[MAXN],n,now,ans,sum[MAXN],cnt,tot,fa[MAXN];
int root(int x)
{
	if(x!=fa[x]) fa[x]=root(fa[x]);
	return fa[x];
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	for(int i=1;i<=n;i++)
		scanf("%d",&b[i]);
	for(int i=1;i<=n;i++)
		now^=a[i];
	ma[now]++;
	for(int i=1;i<=n;i++)
	{
		if(b[i]==now&&a[i]!=now) tot++;
		ma[a[i]]++;
		ma[b[i]]--;
		if(a[i]!=b[i]) ans++;
	}
	int ok=-1;
	for(map<int,int>::iterator it=ma.begin();it!=ma.end();it++)
	{
		if(it->second)
		{
			if(ok!=-1)
			{
				printf("-1");
				return 0;
			}
			ok=it->first;
		}
	}
	if(ans&&tot) ans--;
	for(map<int,int>::iterator it=ma.begin();it!=ma.end();it++)
		it->second=++cnt;
	for(int i=1;i<=cnt;i++)
		fa[i]=i;
	for(int i=1;i<=n;i++)
	{
		if(a[i]!=b[i])
		{
			fa[root(ma[a[i]])]=root(ma[b[i]]);
		}
	}
	for(int i=1;i<=cnt;i++)
		sum[root(i)]++;
	for(int i=1;i<=cnt;i++)
		if(sum[i]>1) ans++;
	printf("%d",ans);
}
### Prompt

Construct a cpp code solution to the problem outlined:

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<cstdio>
#include<map>
#include<algorithm>
#define MAXN 100006
using namespace std;
map<int,int> ma;
int a[MAXN],b[MAXN],n,now,ans,sum[MAXN],cnt,tot,fa[MAXN];
int root(int x)
{
	if(x!=fa[x]) fa[x]=root(fa[x]);
	return fa[x];
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	for(int i=1;i<=n;i++)
		scanf("%d",&b[i]);
	for(int i=1;i<=n;i++)
		now^=a[i];
	ma[now]++;
	for(int i=1;i<=n;i++)
	{
		if(b[i]==now&&a[i]!=now) tot++;
		ma[a[i]]++;
		ma[b[i]]--;
		if(a[i]!=b[i]) ans++;
	}
	int ok=-1;
	for(map<int,int>::iterator it=ma.begin();it!=ma.end();it++)
	{
		if(it->second)
		{
			if(ok!=-1)
			{
				printf("-1");
				return 0;
			}
			ok=it->first;
		}
	}
	if(ans&&tot) ans--;
	for(map<int,int>::iterator it=ma.begin();it!=ma.end();it++)
		it->second=++cnt;
	for(int i=1;i<=cnt;i++)
		fa[i]=i;
	for(int i=1;i<=n;i++)
	{
		if(a[i]!=b[i])
		{
			fa[root(ma[a[i]])]=root(ma[b[i]]);
		}
	}
	for(int i=1;i<=cnt;i++)
		sum[root(i)]++;
	for(int i=1;i<=cnt;i++)
		if(sum[i]>1) ans++;
	printf("%d",ans);
}
```