#include<cstdio>
#include<map>
#include<cstring>
#include<algorithm>
#define LL long long
#define fo(i,j,k) for(int i=j;i<=k;i++)
#define fd(i,j,k) for(int i=j;i>=k;i--)
#define fr(i,j) for(int i=beg[j];i;i=nex[i])
using namespace std;
int const mn=1e5+3;
int n,a[mn],b[mn],cnt[mn],par[mn];
map<int,int>re;
int get(int x){
	return (par[x]==x)?x:par[x]=get(par[x]);
}
int main(){
	scanf("%d",&n);
	int tmp=0;
	fo(i,1,n)scanf("%d",&a[i]),a[n+1]^=a[i];
	fo(i,1,n)scanf("%d",&b[i]),b[n+1]^=b[i];
	n++;
	fo(i,1,n){
		if(!re[a[i]])re[a[i]]=++tmp;
		if(!re[b[i]])re[b[i]]=++tmp;
		cnt[re[a[i]]]++;
		cnt[re[b[i]]]--;
	}
	fo(i,1,tmp)if(cnt[i]){printf("-1");return 0;}
	re.clear();
	tmp=0;
	fo(i,1,n)if((a[i]!=b[i])||(i==n)){
		if(!re[a[i]])re[a[i]]=++tmp;
		if(!re[b[i]])re[b[i]]=++tmp;
	}
	int ans=0;
	fo(i,1,tmp)par[i]=i;
	fo(i,1,n)if(a[i]!=b[i]){
		ans+=i!=n;
		par[get(re[a[i]])]=get(re[b[i]]);
	}
	fo(i,1,tmp)ans+=par[i]==i;
	printf("%d",ans-1);
	return 0;
}

### Prompt

Create a solution in Cpp for 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<map>
#include<cstring>
#include<algorithm>
#define LL long long
#define fo(i,j,k) for(int i=j;i<=k;i++)
#define fd(i,j,k) for(int i=j;i>=k;i--)
#define fr(i,j) for(int i=beg[j];i;i=nex[i])
using namespace std;
int const mn=1e5+3;
int n,a[mn],b[mn],cnt[mn],par[mn];
map<int,int>re;
int get(int x){
	return (par[x]==x)?x:par[x]=get(par[x]);
}
int main(){
	scanf("%d",&n);
	int tmp=0;
	fo(i,1,n)scanf("%d",&a[i]),a[n+1]^=a[i];
	fo(i,1,n)scanf("%d",&b[i]),b[n+1]^=b[i];
	n++;
	fo(i,1,n){
		if(!re[a[i]])re[a[i]]=++tmp;
		if(!re[b[i]])re[b[i]]=++tmp;
		cnt[re[a[i]]]++;
		cnt[re[b[i]]]--;
	}
	fo(i,1,tmp)if(cnt[i]){printf("-1");return 0;}
	re.clear();
	tmp=0;
	fo(i,1,n)if((a[i]!=b[i])||(i==n)){
		if(!re[a[i]])re[a[i]]=++tmp;
		if(!re[b[i]])re[b[i]]=++tmp;
	}
	int ans=0;
	fo(i,1,tmp)par[i]=i;
	fo(i,1,n)if(a[i]!=b[i]){
		ans+=i!=n;
		par[get(re[a[i]])]=get(re[b[i]]);
	}
	fo(i,1,tmp)ans+=par[i]==i;
	printf("%d",ans-1);
	return 0;
}

```