#include<bits/stdc++.h>
#define N 100005
using namespace std;
int buf[N],bcnt,*x[N];
bool cmp(int *a,int *b){
	return (*a)<(*b);
}
void lsh(int *a,int sz){
	for(int i=1;i<=sz;i++) x[i]=&a[i];
	bcnt=0;sort(x+1,x+sz+1,cmp);
	buf[1]=++bcnt;
	for(int i=2;i<=sz;i++) buf[i]=*x[i]==*x[i-1] ? buf[i-1]:++bcnt;
	for(int i=1;i<=sz;i++) *x[i]=buf[i];
}

int n;
int bcj[N];
int ff(int x){
	return bcj[x]==x? x:bcj[x]=ff(bcj[x]);
}
void connect(int x,int y){
	if(ff(x)==ff(y)) return;
	bcj[ff(x)]=ff(y);
}

int a[N],b[N];
int c[N],d[N],m;
int main(){
	scanf("%d",&n);int ta=0,tb=0;
	for(int i=1;i<=n;i++) scanf("%d",&a[i]),ta^=a[i];
	for(int i=1;i<=n;i++) scanf("%d",&b[i]),tb^=b[i];
	n++;
	a[n]=ta;b[n]=tb;
	for(int i=1;i<=n;i++)
		if(a[i]!=b[i])
			c[++m]=a[i],d[m]=b[i];
	int fl1=0,fl2=0;
	if(a[n]==b[n]) fl1=1;
	for(int i=1;i<=m;i++) if(c[i]==a[n]) fl2=1;
	sort(a+1,a+n+1);
	sort(b+1,b+n+1);
	for(int i=1;i<=n;i++) if(a[i]!=b[i]) return puts("-1"),0;
	
	int ans=0;
	if(fl1==1&&fl2==1) ans=-1;
	if(fl1==0) ans-=2;
	ans+=m;
	
	lsh(c,m);
	lsh(d,m);
	for(int i=1;i<=m;i++) bcj[i]=i;
	for(int i=1;i<=m;i++) connect(c[i],d[i]);
	for(int i=1;i<=bcnt;i++) if(ff(i)==i) ans++;
	
	cout<<ans<<endl;
	return 0;
} 
### 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<bits/stdc++.h>
#define N 100005
using namespace std;
int buf[N],bcnt,*x[N];
bool cmp(int *a,int *b){
	return (*a)<(*b);
}
void lsh(int *a,int sz){
	for(int i=1;i<=sz;i++) x[i]=&a[i];
	bcnt=0;sort(x+1,x+sz+1,cmp);
	buf[1]=++bcnt;
	for(int i=2;i<=sz;i++) buf[i]=*x[i]==*x[i-1] ? buf[i-1]:++bcnt;
	for(int i=1;i<=sz;i++) *x[i]=buf[i];
}

int n;
int bcj[N];
int ff(int x){
	return bcj[x]==x? x:bcj[x]=ff(bcj[x]);
}
void connect(int x,int y){
	if(ff(x)==ff(y)) return;
	bcj[ff(x)]=ff(y);
}

int a[N],b[N];
int c[N],d[N],m;
int main(){
	scanf("%d",&n);int ta=0,tb=0;
	for(int i=1;i<=n;i++) scanf("%d",&a[i]),ta^=a[i];
	for(int i=1;i<=n;i++) scanf("%d",&b[i]),tb^=b[i];
	n++;
	a[n]=ta;b[n]=tb;
	for(int i=1;i<=n;i++)
		if(a[i]!=b[i])
			c[++m]=a[i],d[m]=b[i];
	int fl1=0,fl2=0;
	if(a[n]==b[n]) fl1=1;
	for(int i=1;i<=m;i++) if(c[i]==a[n]) fl2=1;
	sort(a+1,a+n+1);
	sort(b+1,b+n+1);
	for(int i=1;i<=n;i++) if(a[i]!=b[i]) return puts("-1"),0;
	
	int ans=0;
	if(fl1==1&&fl2==1) ans=-1;
	if(fl1==0) ans-=2;
	ans+=m;
	
	lsh(c,m);
	lsh(d,m);
	for(int i=1;i<=m;i++) bcj[i]=i;
	for(int i=1;i<=m;i++) connect(c[i],d[i]);
	for(int i=1;i<=bcnt;i++) if(ff(i)==i) ans++;
	
	cout<<ans<<endl;
	return 0;
} 
```