#include<iostream>
#include<cstdio>
#include<map>
#define MN 200000
using namespace std;
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
map<int,int>mp;
int now=0,n,a[MN+5],b[MN+5],cnt,head[MN+5],in[MN+5],out[MN+5],num=0,size[MN+5],s[MN+5];
inline int getfa(int x){return s[x]?s[x]=getfa(s[x]):x;}
inline void ins(int f,int t)
{
	if(getfa(f)!=getfa(t))
		size[getfa(t)]+=size[getfa(f)],s[getfa(f)]=getfa(t);
	++in[t];++out[f];++cnt;
}
int main()
{
	n=read();
	for(int i=1;i<=n;++i) 
	{
		a[i]=read(),now^=a[i];
		if(!mp[a[i]]) mp[a[i]]=++num;
	}
	for(int i=1;i<=n;++i)
	{
		b[i]=read();
		if(!mp[b[i]]) mp[b[i]]=++num;	
	}
	if(!mp[now]) mp[now]=++num;++in[mp[now]];
	for(int i=1;i<=num;++i) size[i]=1;
	for(int i=1;i<=n;++i) if(a[i]!=b[i]) ins(mp[b[i]],mp[a[i]]);
	int difnum=0;
	for(int i=1;i<=num;++i) 
		if(out[i]>in[i]||in[i]+1<out[i]) return 0*puts("-1");
		else if(out[i]!=in[i]) ++difnum;
	if(difnum>1) return 0*puts("-1");
	for(int i=1;i<=num;++i) if(getfa(i)==i)
	{	
		if(size[i]==1) cnt+=(mp[now]==i);
		else ++cnt;
	}
	printf("%d",cnt-1);
	return 0;
}


### Prompt

Please provide a Cpp coded solution 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<iostream>
#include<cstdio>
#include<map>
#define MN 200000
using namespace std;
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
map<int,int>mp;
int now=0,n,a[MN+5],b[MN+5],cnt,head[MN+5],in[MN+5],out[MN+5],num=0,size[MN+5],s[MN+5];
inline int getfa(int x){return s[x]?s[x]=getfa(s[x]):x;}
inline void ins(int f,int t)
{
	if(getfa(f)!=getfa(t))
		size[getfa(t)]+=size[getfa(f)],s[getfa(f)]=getfa(t);
	++in[t];++out[f];++cnt;
}
int main()
{
	n=read();
	for(int i=1;i<=n;++i) 
	{
		a[i]=read(),now^=a[i];
		if(!mp[a[i]]) mp[a[i]]=++num;
	}
	for(int i=1;i<=n;++i)
	{
		b[i]=read();
		if(!mp[b[i]]) mp[b[i]]=++num;	
	}
	if(!mp[now]) mp[now]=++num;++in[mp[now]];
	for(int i=1;i<=num;++i) size[i]=1;
	for(int i=1;i<=n;++i) if(a[i]!=b[i]) ins(mp[b[i]],mp[a[i]]);
	int difnum=0;
	for(int i=1;i<=num;++i) 
		if(out[i]>in[i]||in[i]+1<out[i]) return 0*puts("-1");
		else if(out[i]!=in[i]) ++difnum;
	if(difnum>1) return 0*puts("-1");
	for(int i=1;i<=num;++i) if(getfa(i)==i)
	{	
		if(size[i]==1) cnt+=(mp[now]==i);
		else ++cnt;
	}
	printf("%d",cnt-1);
	return 0;
}


```