#include<bits/stdc++.h>
#define LL long long
const int N=200005;
using namespace std;

int n,a[N],b[N],num[N],tot=0;
int fa[N],sz[N],ans=0,chk[N];
map<int,int>vis;

int getfa(int x){
	if(fa[x]!=x)fa[x]=getfa(fa[x]);
	return fa[x];
}

void add(int x,int y){
	int X=getfa(x),Y=getfa(y);
	if(X==Y)return;
	fa[X]=Y;chk[Y]=1;
}

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d",&a[i]);
		if(!vis[a[i]])vis[a[i]]=++tot;
		a[n+1]^=a[i];
		num[vis[a[i]]]++;
	}
	if(!vis[a[n+1]])vis[a[n+1]]=++tot;
	num[vis[a[n+1]]]++;
	for(int i=1;i<=n;i++){
		scanf("%d",&b[i]);
		if(!vis[b[i]]||num[vis[b[i]]]<=0)ans=-1;
		b[n+1]^=b[i];
		num[vis[b[i]]]--;
	}
	if(!vis[b[n+1]]||num[vis[b[n+1]]]<=0)ans=-1;
	if(ans==-1){
		printf("-1\n");
		return 0;
	}
	for(int i=1;i<=tot;i++)fa[i]=i;
	for(int i=1;i<=n;i++){
		if(a[i]==b[i])continue;
		ans++;
		add(vis[a[i]],vis[b[i]]);
	}
	if(vis[a[n+1]]!=vis[b[n+1]])add(vis[a[n+1]],vis[b[n+1]]);
	for(int i=1;i<=tot;i++)fa[i]=getfa(i);
	for(int i=1;i<=tot;i++){
		if(fa[i]==i&&chk[i])ans++;
		if(i==vis[a[n+1]]){
			if(fa[i]!=i||(fa[i]==i&&chk[i]))ans--;
		}
	}
	printf("%d\n",ans);
}
### Prompt

Your task is to create 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 LL long long
const int N=200005;
using namespace std;

int n,a[N],b[N],num[N],tot=0;
int fa[N],sz[N],ans=0,chk[N];
map<int,int>vis;

int getfa(int x){
	if(fa[x]!=x)fa[x]=getfa(fa[x]);
	return fa[x];
}

void add(int x,int y){
	int X=getfa(x),Y=getfa(y);
	if(X==Y)return;
	fa[X]=Y;chk[Y]=1;
}

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d",&a[i]);
		if(!vis[a[i]])vis[a[i]]=++tot;
		a[n+1]^=a[i];
		num[vis[a[i]]]++;
	}
	if(!vis[a[n+1]])vis[a[n+1]]=++tot;
	num[vis[a[n+1]]]++;
	for(int i=1;i<=n;i++){
		scanf("%d",&b[i]);
		if(!vis[b[i]]||num[vis[b[i]]]<=0)ans=-1;
		b[n+1]^=b[i];
		num[vis[b[i]]]--;
	}
	if(!vis[b[n+1]]||num[vis[b[n+1]]]<=0)ans=-1;
	if(ans==-1){
		printf("-1\n");
		return 0;
	}
	for(int i=1;i<=tot;i++)fa[i]=i;
	for(int i=1;i<=n;i++){
		if(a[i]==b[i])continue;
		ans++;
		add(vis[a[i]],vis[b[i]]);
	}
	if(vis[a[n+1]]!=vis[b[n+1]])add(vis[a[n+1]],vis[b[n+1]]);
	for(int i=1;i<=tot;i++)fa[i]=getfa(i);
	for(int i=1;i<=tot;i++){
		if(fa[i]==i&&chk[i])ans++;
		if(i==vis[a[n+1]]){
			if(fa[i]!=i||(fa[i]==i&&chk[i]))ans--;
		}
	}
	printf("%d\n",ans);
}
```