#include<bits/stdc++.h>
#define For(i,l,r) for(int i=l;i<=r;i++)
#define ll long long
using namespace std;
const int M=1e5+5;
int n,a[M],b[M],x[M],y[M],vis[M],cnt,ans;
map<int,int>d;
vector<int>Q[M];
inline int read(){
	int f=1,sum=0;
	char ch=getchar();
	while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
	while(isdigit(ch)){sum=(sum<<1)+(sum<<3)+(ch^48);ch=getchar();}
	return f*sum; 
}
inline void dfs(int x){
	vis[x]=1;
	For(i,0,Q[x].size()-1){
		if(!vis[Q[x][i]]){
			dfs(Q[x][i]);
		}
	}
}
int main(){
	n=read();
	For(i,1,n){
		a[i]=read();a[n+1]^=a[i];
	    if(!d[a[i]]) d[a[i]]=++cnt;
	}
	if(!d[a[n+1]]) d[a[n+1]]=++cnt;
	For(i,1,n){
		b[i]=read();b[n+1]^=b[i];
		if(!d[b[i]]) d[b[i]]=++cnt;
	}
	if(!d[b[n+1]]) d[b[n+1]]=++cnt;
	memcpy(x,a,sizeof(x));memcpy(y,b,sizeof(y));
	sort(x+1,x+n+2);sort(y+1,y+n+2);
	For(i,1,n) if(x[i]!=y[i]){printf("-1");return 0;}
	For(i,1,n+1) a[i]=d[a[i]],b[i]=d[b[i]];
	For(i,1,n) if(a[i]!=b[i]) ans++,Q[a[i]].push_back(b[i]);
	if(a[n+1]!=b[n+1]) Q[a[n+1]].push_back(b[n+1]); 
	For(i,1,cnt-1) if(Q[i].size()&&!vis[i]) dfs(i),ans++;
	if(!vis[cnt]) ans++;
	printf("%d",ans-1);
	return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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 For(i,l,r) for(int i=l;i<=r;i++)
#define ll long long
using namespace std;
const int M=1e5+5;
int n,a[M],b[M],x[M],y[M],vis[M],cnt,ans;
map<int,int>d;
vector<int>Q[M];
inline int read(){
	int f=1,sum=0;
	char ch=getchar();
	while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
	while(isdigit(ch)){sum=(sum<<1)+(sum<<3)+(ch^48);ch=getchar();}
	return f*sum; 
}
inline void dfs(int x){
	vis[x]=1;
	For(i,0,Q[x].size()-1){
		if(!vis[Q[x][i]]){
			dfs(Q[x][i]);
		}
	}
}
int main(){
	n=read();
	For(i,1,n){
		a[i]=read();a[n+1]^=a[i];
	    if(!d[a[i]]) d[a[i]]=++cnt;
	}
	if(!d[a[n+1]]) d[a[n+1]]=++cnt;
	For(i,1,n){
		b[i]=read();b[n+1]^=b[i];
		if(!d[b[i]]) d[b[i]]=++cnt;
	}
	if(!d[b[n+1]]) d[b[n+1]]=++cnt;
	memcpy(x,a,sizeof(x));memcpy(y,b,sizeof(y));
	sort(x+1,x+n+2);sort(y+1,y+n+2);
	For(i,1,n) if(x[i]!=y[i]){printf("-1");return 0;}
	For(i,1,n+1) a[i]=d[a[i]],b[i]=d[b[i]];
	For(i,1,n) if(a[i]!=b[i]) ans++,Q[a[i]].push_back(b[i]);
	if(a[n+1]!=b[n+1]) Q[a[n+1]].push_back(b[n+1]); 
	For(i,1,cnt-1) if(Q[i].size()&&!vis[i]) dfs(i),ans++;
	if(!vis[cnt]) ans++;
	printf("%d",ans-1);
	return 0;
}
```