#include<bits/stdc++.h>
#define maxn 100010
#define pb push_back
using namespace std;

int a[maxn],b[maxn],c[maxn],d[maxn],fa[maxn],sz[maxn],n,cnt=0,cnt2=0;
bool vis[maxn];
vector<int> v;
int find(int x){
	return fa[x]==x?x:fa[x]=find(fa[x]);
}
int main(){
	ios::sync_with_stdio(0);
	cin>>n;
	for(int i=1;i<=n;i++)
		cin>>a[i],a[n+1]^=a[i],v.pb(a[i]);
	v.pb(a[n+1]);
	for(int i=1;i<=n;i++)
		cin>>b[i],b[n+1]^=b[i];
	memcpy(c,a,sizeof(a));
	memcpy(d,b,sizeof(b));
	sort(c+1,c+n+2),sort(d+1,d+n+2);
	for(int i=1;i<=n+1;i++){
		if(c[i]!=d[i])
			return cout<<"-1"<<endl,0;
	}
	sort(v.begin(),v.end());
	v.erase(unique(v.begin(),v.end()),v.end());
	for(int i=0;i<v.size();i++)
		fa[i]=i,sz[i]=1;
	for(int i=1;i<=n+1;i++){
		int x=lower_bound(v.begin(),v.end(),a[i])-v.begin();
		int y=lower_bound(v.begin(),v.end(),b[i])-v.begin();
		if(a[i]!=b[i]||i==n+1)
			cnt+=i<=n,vis[x]=vis[y]=1;
		if(find(x)!=find(y))
			fa[find(x)]=find(y);
	}
	if(!cnt)
		return cout<<"0"<<endl,0;
	for(int i=0;i<v.size();i++)
		cnt+=vis[i]&&find(i)==i;
	cout<<cnt-1<<endl;
	return 0;
}
### Prompt

Your challenge is to write 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 maxn 100010
#define pb push_back
using namespace std;

int a[maxn],b[maxn],c[maxn],d[maxn],fa[maxn],sz[maxn],n,cnt=0,cnt2=0;
bool vis[maxn];
vector<int> v;
int find(int x){
	return fa[x]==x?x:fa[x]=find(fa[x]);
}
int main(){
	ios::sync_with_stdio(0);
	cin>>n;
	for(int i=1;i<=n;i++)
		cin>>a[i],a[n+1]^=a[i],v.pb(a[i]);
	v.pb(a[n+1]);
	for(int i=1;i<=n;i++)
		cin>>b[i],b[n+1]^=b[i];
	memcpy(c,a,sizeof(a));
	memcpy(d,b,sizeof(b));
	sort(c+1,c+n+2),sort(d+1,d+n+2);
	for(int i=1;i<=n+1;i++){
		if(c[i]!=d[i])
			return cout<<"-1"<<endl,0;
	}
	sort(v.begin(),v.end());
	v.erase(unique(v.begin(),v.end()),v.end());
	for(int i=0;i<v.size();i++)
		fa[i]=i,sz[i]=1;
	for(int i=1;i<=n+1;i++){
		int x=lower_bound(v.begin(),v.end(),a[i])-v.begin();
		int y=lower_bound(v.begin(),v.end(),b[i])-v.begin();
		if(a[i]!=b[i]||i==n+1)
			cnt+=i<=n,vis[x]=vis[y]=1;
		if(find(x)!=find(y))
			fa[find(x)]=find(y);
	}
	if(!cnt)
		return cout<<"0"<<endl,0;
	for(int i=0;i<v.size();i++)
		cnt+=vis[i]&&find(i)==i;
	cout<<cnt-1<<endl;
	return 0;
}
```