//#pragma comment(linker, "/STACK:1024000000,1024000000")
//#pragma GCC optimize (2)
//#pragma G++ optimize (2)
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 100010
int n,a[N],b[N],as[N],bs[N],al,bl,f[N],ans;
int find(int x){ return x==f[x] ? x : f[x]=find(f[x]); }
int main(){
	//freopen("duliu_3.in","r",stdin);
	//freopen("duliu.out","w",stdout);
	scanf("%d",&n);
	for(int i=1;i<=n;++i){ scanf("%d",&a[i]); as[i]=a[i]; a[n+1]^=a[i]; }
	for(int i=1;i<=n;++i){ scanf("%d",&b[i]); bs[i]=b[i]; b[n+1]^=b[i]; }
	as[n+1]=a[n+1]; bs[n+1]=b[n+1];
	sort(as+1,as+n+2); 
	sort(bs+1,bs+n+2); 
	for(int i=1;i<=n+1;++i)if(as[i]^bs[i]){puts("-1");return 0;}
	al=unique(as+1,as+n+2)-(as+1);
	bl=unique(bs+1,bs+n+2)-(bs+1);
	if(al^bl){ puts("-1"); return 0; }
	for(int i=1;i<=n+1;++i){
		a[i]=lower_bound(as+1,as+1+al,a[i])-as;
		b[i]=lower_bound(bs+1,bs+1+bl,b[i])-bs;
	}
	for(int i=1;i<=n;++i) if(a[i]^b[i]){ f[a[i]]=a[i]; f[b[i]]=b[i]; }
	f[a[n+1]]=a[n+1];f[b[n+1]]=b[n+1];
	for(int fa,fb,i=1;i<=n+1;++i) 
		if(a[i]^b[i]){
			if(i<=n) ++ans;
			fa=find(a[i]); 
			fb=find(b[i]);
			f[fa]=fb;
		}
	for(int fi,i=1;i<=al;++i){
		fi=find(i); 
		if(fi==i)	++ans;
	}
	ans-1>0 ? ans-=1 : ans=0;
	printf("%d",ans);
	return 0;
}
### Prompt

Your task is to create 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
//#pragma comment(linker, "/STACK:1024000000,1024000000")
//#pragma GCC optimize (2)
//#pragma G++ optimize (2)
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 100010
int n,a[N],b[N],as[N],bs[N],al,bl,f[N],ans;
int find(int x){ return x==f[x] ? x : f[x]=find(f[x]); }
int main(){
	//freopen("duliu_3.in","r",stdin);
	//freopen("duliu.out","w",stdout);
	scanf("%d",&n);
	for(int i=1;i<=n;++i){ scanf("%d",&a[i]); as[i]=a[i]; a[n+1]^=a[i]; }
	for(int i=1;i<=n;++i){ scanf("%d",&b[i]); bs[i]=b[i]; b[n+1]^=b[i]; }
	as[n+1]=a[n+1]; bs[n+1]=b[n+1];
	sort(as+1,as+n+2); 
	sort(bs+1,bs+n+2); 
	for(int i=1;i<=n+1;++i)if(as[i]^bs[i]){puts("-1");return 0;}
	al=unique(as+1,as+n+2)-(as+1);
	bl=unique(bs+1,bs+n+2)-(bs+1);
	if(al^bl){ puts("-1"); return 0; }
	for(int i=1;i<=n+1;++i){
		a[i]=lower_bound(as+1,as+1+al,a[i])-as;
		b[i]=lower_bound(bs+1,bs+1+bl,b[i])-bs;
	}
	for(int i=1;i<=n;++i) if(a[i]^b[i]){ f[a[i]]=a[i]; f[b[i]]=b[i]; }
	f[a[n+1]]=a[n+1];f[b[n+1]]=b[n+1];
	for(int fa,fb,i=1;i<=n+1;++i) 
		if(a[i]^b[i]){
			if(i<=n) ++ans;
			fa=find(a[i]); 
			fb=find(b[i]);
			f[fa]=fb;
		}
	for(int fi,i=1;i<=al;++i){
		fi=find(i); 
		if(fi==i)	++ans;
	}
	ans-1>0 ? ans-=1 : ans=0;
	printf("%d",ans);
	return 0;
}
```