#include<bits/stdc++.h>
#define N 200005
using namespace std;
int now=0,n,cnt,num;
int a[N],b[N],fa[N];
int in[N],out[N];
int sz[N];
map<int,int> mp;
int get(int x){
	return fa[x]==x?x:fa[x]=get(fa[x]);
}
void add(int x,int y){
	in[y]++; out[x]++; cnt++;
	x=get(x); y=get(y);
	if (x!=y) sz[y]+=sz[x],fa[x]=y;
}
int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;i++){
		scanf("%d",&a[i]);
		now^=a[i];
		if (!mp[a[i]]) mp[a[i]]=++num;
	}
	for (int i=1;i<=n;i++){
		scanf("%d",&b[i]);
		if (!mp[b[i]]) mp[b[i]]=++num;
	}
	if (!mp[now]) mp[now]=++num;
	++in[mp[now]];
	for (int i=1;i<=num;i++) sz[i]=1,fa[i]=i;
	for (int i=1;i<=n;i++)
		if (a[i]!=b[i])
			add(mp[b[i]],mp[a[i]]);
	int dif=0;
	for (int i=1;i<=num;i++)
		if (out[i]>in[i]||in[i]+1<out[i])
			return puts("-1"),0;
		else if (out[i]!=in[i]) ++dif;
	if (dif>1) return puts("-2"),0;
	for (int i=1;i<=num;i++)
		if (get(i)==i){
			if (sz[i]==1) cnt+=(mp[now]==i);
			else ++cnt;
		}
	printf("%d\n",cnt-1);
}
### Prompt

Generate 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 200005
using namespace std;
int now=0,n,cnt,num;
int a[N],b[N],fa[N];
int in[N],out[N];
int sz[N];
map<int,int> mp;
int get(int x){
	return fa[x]==x?x:fa[x]=get(fa[x]);
}
void add(int x,int y){
	in[y]++; out[x]++; cnt++;
	x=get(x); y=get(y);
	if (x!=y) sz[y]+=sz[x],fa[x]=y;
}
int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;i++){
		scanf("%d",&a[i]);
		now^=a[i];
		if (!mp[a[i]]) mp[a[i]]=++num;
	}
	for (int i=1;i<=n;i++){
		scanf("%d",&b[i]);
		if (!mp[b[i]]) mp[b[i]]=++num;
	}
	if (!mp[now]) mp[now]=++num;
	++in[mp[now]];
	for (int i=1;i<=num;i++) sz[i]=1,fa[i]=i;
	for (int i=1;i<=n;i++)
		if (a[i]!=b[i])
			add(mp[b[i]],mp[a[i]]);
	int dif=0;
	for (int i=1;i<=num;i++)
		if (out[i]>in[i]||in[i]+1<out[i])
			return puts("-1"),0;
		else if (out[i]!=in[i]) ++dif;
	if (dif>1) return puts("-2"),0;
	for (int i=1;i<=num;i++)
		if (get(i)==i){
			if (sz[i]==1) cnt+=(mp[now]==i);
			else ++cnt;
		}
	printf("%d\n",cnt-1);
}
```