#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#define fo(i,a,b) for(i=a;i<=b;i++)

using namespace std;

const int maxn=1e5+5;

int a[maxn],b[maxn],c[maxn],d[maxn],fa[maxn];
int i,j,k,l,m,n,x,y,ans,p;
map<int,int> f;

int getfather(int x){
	if (fa[x]==x) return x; else fa[x]=getfather(fa[x]);
	return fa[x];
}
int main(){
//	freopen("016d.in","r",stdin);
	scanf("%d",&n);
	l=0;
	fo(i,1,n){
		scanf("%d",&a[i]);
		l^=a[i];
	}
	a[n+1]=l; l=0;
	fo(i,1,n){
		scanf("%d",&b[i]);
		l^=b[i];
	}
	b[++n]=l;
	fo(i,1,n){
		c[i]=a[i]; d[i]=b[i];
	}
	sort(c+1,c+n+1); sort(d+1,d+n+1);
	fo(i,1,n-1) if (c[i]!=d[i]) break;
	if (i<n){
		printf("-1\n");
		return 0;
	}
	fo(i,1,n){
		if (a[i]!=b[i] && i!=n) ans++;
		if (i==n && ans==0){
			printf("0\n"); return 0;
		}
		if (a[i]!=b[i] || i==n){
			if (!f[a[i]]) f[a[i]]=++p;
			if (!f[b[i]]) f[b[i]]=++p;
		}
	}
	fo(i,1,p) fa[i]=i;
	fo(i,1,n) if (a[i]!=b[i]) fa[getfather(f[a[i]])]=getfather(f[b[i]]);
	fo(i,1,p) if (getfather(i)==i) ans++;
	printf("%d\n",ans-1);
	return 0;
}
### Prompt

In CPP, your task is to solve 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<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#define fo(i,a,b) for(i=a;i<=b;i++)

using namespace std;

const int maxn=1e5+5;

int a[maxn],b[maxn],c[maxn],d[maxn],fa[maxn];
int i,j,k,l,m,n,x,y,ans,p;
map<int,int> f;

int getfather(int x){
	if (fa[x]==x) return x; else fa[x]=getfather(fa[x]);
	return fa[x];
}
int main(){
//	freopen("016d.in","r",stdin);
	scanf("%d",&n);
	l=0;
	fo(i,1,n){
		scanf("%d",&a[i]);
		l^=a[i];
	}
	a[n+1]=l; l=0;
	fo(i,1,n){
		scanf("%d",&b[i]);
		l^=b[i];
	}
	b[++n]=l;
	fo(i,1,n){
		c[i]=a[i]; d[i]=b[i];
	}
	sort(c+1,c+n+1); sort(d+1,d+n+1);
	fo(i,1,n-1) if (c[i]!=d[i]) break;
	if (i<n){
		printf("-1\n");
		return 0;
	}
	fo(i,1,n){
		if (a[i]!=b[i] && i!=n) ans++;
		if (i==n && ans==0){
			printf("0\n"); return 0;
		}
		if (a[i]!=b[i] || i==n){
			if (!f[a[i]]) f[a[i]]=++p;
			if (!f[b[i]]) f[b[i]]=++p;
		}
	}
	fo(i,1,p) fa[i]=i;
	fo(i,1,n) if (a[i]!=b[i]) fa[getfather(f[a[i]])]=getfather(f[b[i]]);
	fo(i,1,p) if (getfather(i)==i) ans++;
	printf("%d\n",ans-1);
	return 0;
}
```