#include<bits/stdc++.h>

using namespace std;

int n,a[100010],a1[100010],b[100010],b1[100010];
int ans=-1,fa[100010],h[19260827],p[19260827],cnt;

inline int hah(int x)
{
	int y=x%19260817;
	while(h[y]>=0&&h[y]!=x)y=(y+1)%19260817;
	return y;
}

inline int f(int x)
{
	if(fa[x]!=x)fa[x]=f(fa[x]);
	return fa[x];
}

int main()
{
	scanf("%d",&n);
	for(int i=1; i<=n; i++)
	{
		scanf("%d",&a[i]);
		a1[i]=a[i];
		a[n+1]^=a[i];
	}
	a1[n+1]=a[n+1];
	for(int i=1; i<=n; i++)
	{
		scanf("%d",&b[i]);
		b1[i]=b[i];
		b[n+1]^=b[i];
	}
	b1[n+1]=b[n+1];
	sort(a1+1,a1+n+2);
	sort(b1+1,b1+n+2);
	for(int i=1; i<=n+1; i++)
	{
		if(a1[i]^b1[i])
		{
			printf("-1");
			return 0;
		}
	}
	memset(h,-1,sizeof(h));
	for(int i=1; i<=n+1; i++)
	{
		if(i==n+1||a[i]^b[i])
		{
			if(i<=n)ans++;
			int k=hah(a[i]);
			if(h[k]==-1)h[k]=a[i],p[k]=++cnt;
			k=hah(b[i]);
			if(h[k]==-1)h[k]=b[i],p[k]=++cnt;
		}
	}
	for(int i=1; i<=cnt; i++)fa[i]=i;
	for(int i=1; i<=n+1; i++)
	{
		if(a[i]^b[i])
		{
			fa[f(p[hah(a[i])])]=f(p[hah(b[i])]);
		}
	}
	for(int i=1; i<=cnt; i++)if(fa[i]==i)ans++;
	printf("%d",ans);
	return 0;
}
### Prompt

Please create a solution in cpp 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;

int n,a[100010],a1[100010],b[100010],b1[100010];
int ans=-1,fa[100010],h[19260827],p[19260827],cnt;

inline int hah(int x)
{
	int y=x%19260817;
	while(h[y]>=0&&h[y]!=x)y=(y+1)%19260817;
	return y;
}

inline int f(int x)
{
	if(fa[x]!=x)fa[x]=f(fa[x]);
	return fa[x];
}

int main()
{
	scanf("%d",&n);
	for(int i=1; i<=n; i++)
	{
		scanf("%d",&a[i]);
		a1[i]=a[i];
		a[n+1]^=a[i];
	}
	a1[n+1]=a[n+1];
	for(int i=1; i<=n; i++)
	{
		scanf("%d",&b[i]);
		b1[i]=b[i];
		b[n+1]^=b[i];
	}
	b1[n+1]=b[n+1];
	sort(a1+1,a1+n+2);
	sort(b1+1,b1+n+2);
	for(int i=1; i<=n+1; i++)
	{
		if(a1[i]^b1[i])
		{
			printf("-1");
			return 0;
		}
	}
	memset(h,-1,sizeof(h));
	for(int i=1; i<=n+1; i++)
	{
		if(i==n+1||a[i]^b[i])
		{
			if(i<=n)ans++;
			int k=hah(a[i]);
			if(h[k]==-1)h[k]=a[i],p[k]=++cnt;
			k=hah(b[i]);
			if(h[k]==-1)h[k]=b[i],p[k]=++cnt;
		}
	}
	for(int i=1; i<=cnt; i++)fa[i]=i;
	for(int i=1; i<=n+1; i++)
	{
		if(a[i]^b[i])
		{
			fa[f(p[hah(a[i])])]=f(p[hah(b[i])]);
		}
	}
	for(int i=1; i<=cnt; i++)if(fa[i]==i)ans++;
	printf("%d",ans);
	return 0;
}
```