#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define N 101000
using namespace std;
int a[N],b[N],n,m=0,f[N],ans=0,q[N],w[N];
map<int,int> c;
int gf(int x){return f[x]==0?x:f[x]=gf(f[x]);}
void pedg(int x,int y)
{
	int jy=c[x];
	if(jy==0) c[x]=++m;
	jy=c[y];
	if(jy==0) c[y]=++m;
	x=c[x],y=c[y];
	x=gf(x);y=gf(y);
	if(x!=y) f[x]=y;
}
int main()
{
	scanf("%d",&n);
	fo(i,1,n) scanf("%d",&a[i]),a[n+1]^=a[i],q[i]=a[i];
	fo(i,1,n) scanf("%d",&b[i]),b[n+1]^=b[i],w[i]=b[i];
	q[n+1]=a[n+1];w[n+1]=b[n+1];
	sort(q+1,q+n+2);sort(w+1,w+n+2);
	int jy=0;
	fo(i,1,n) if(a[i]!=b[i]&&b[i]==a[n+1]) jy=1;
	fo(i,1,n+1) if(q[i]!=w[i]) return printf("-1\n"),0;
	fo(i,1,n+1) if(a[i]!=b[i]) pedg(a[i],b[i]),ans+=(i<=n);
	fo(i,1,m) ans+=(f[i]==0);
	printf("%d\n",ans-jy);
}
### Prompt

Please formulate 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
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define N 101000
using namespace std;
int a[N],b[N],n,m=0,f[N],ans=0,q[N],w[N];
map<int,int> c;
int gf(int x){return f[x]==0?x:f[x]=gf(f[x]);}
void pedg(int x,int y)
{
	int jy=c[x];
	if(jy==0) c[x]=++m;
	jy=c[y];
	if(jy==0) c[y]=++m;
	x=c[x],y=c[y];
	x=gf(x);y=gf(y);
	if(x!=y) f[x]=y;
}
int main()
{
	scanf("%d",&n);
	fo(i,1,n) scanf("%d",&a[i]),a[n+1]^=a[i],q[i]=a[i];
	fo(i,1,n) scanf("%d",&b[i]),b[n+1]^=b[i],w[i]=b[i];
	q[n+1]=a[n+1];w[n+1]=b[n+1];
	sort(q+1,q+n+2);sort(w+1,w+n+2);
	int jy=0;
	fo(i,1,n) if(a[i]!=b[i]&&b[i]==a[n+1]) jy=1;
	fo(i,1,n+1) if(q[i]!=w[i]) return printf("-1\n"),0;
	fo(i,1,n+1) if(a[i]!=b[i]) pedg(a[i],b[i]),ans+=(i<=n);
	fo(i,1,m) ans+=(f[i]==0);
	printf("%d\n",ans-jy);
}
```