#include<iostream>
#include<cstdio>
#include<set>
#include<map>
using namespace std;
#define MAX 100100
inline int read()
{
	int x=0;bool t=false;char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
	if(ch=='-')t=true,ch=getchar();
	while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
	return t?-x:x;
}
map<int,int> M;
multiset<int> S;
int f[MAX],tot;
int getf(int x){return x==f[x]?x:f[x]=getf(f[x]);}
int ID(int x){return M[x]?M[x]:M[x]=++tot;}
int n,a[MAX],b[MAX],ans=0;
int main()
{
	n=read();
	for(int i=1;i<=n;++i)a[i]=read(),S.insert(a[i]),a[n+1]^=a[i];
	for(int i=1;i<=n;++i)b[i]=read(),b[n+1]^=b[i];
	S.insert(a[n+1]);
	for(int i=1;i<=n;++i)
		if(S.find(b[i])==S.end()){puts("-1");return 0;}
		else S.erase(S.find(b[i]));
	for(int i=1;i<=n+1;++i)f[i]=i;
	for(int i=1;i<=n;++i)
		if(a[i]!=b[i])
		{
			int u=ID(a[i]),v=ID(b[i]);
			f[getf(u)]=getf(v);++ans;
		}
	if(!ans){puts("0");return 0;}
	for(int i=1;i<=tot;++i)if(getf(i)==i)++ans;
	bool fl=true;
	for(int i=1;i<=n;++i)if(b[i]==a[n+1])fl=false;
	ans+=fl;ans-=1;
	printf("%d\n",ans);
	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<iostream>
#include<cstdio>
#include<set>
#include<map>
using namespace std;
#define MAX 100100
inline int read()
{
	int x=0;bool t=false;char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
	if(ch=='-')t=true,ch=getchar();
	while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
	return t?-x:x;
}
map<int,int> M;
multiset<int> S;
int f[MAX],tot;
int getf(int x){return x==f[x]?x:f[x]=getf(f[x]);}
int ID(int x){return M[x]?M[x]:M[x]=++tot;}
int n,a[MAX],b[MAX],ans=0;
int main()
{
	n=read();
	for(int i=1;i<=n;++i)a[i]=read(),S.insert(a[i]),a[n+1]^=a[i];
	for(int i=1;i<=n;++i)b[i]=read(),b[n+1]^=b[i];
	S.insert(a[n+1]);
	for(int i=1;i<=n;++i)
		if(S.find(b[i])==S.end()){puts("-1");return 0;}
		else S.erase(S.find(b[i]));
	for(int i=1;i<=n+1;++i)f[i]=i;
	for(int i=1;i<=n;++i)
		if(a[i]!=b[i])
		{
			int u=ID(a[i]),v=ID(b[i]);
			f[getf(u)]=getf(v);++ans;
		}
	if(!ans){puts("0");return 0;}
	for(int i=1;i<=tot;++i)if(getf(i)==i)++ans;
	bool fl=true;
	for(int i=1;i<=n;++i)if(b[i]==a[n+1])fl=false;
	ans+=fl;ans-=1;
	printf("%d\n",ans);
	return 0;
}

```