#include<bits/stdc++.h>
using namespace std;
typedef vector<int>VI;
const int N=1e5+5;
int a[N],b[N],n,i,ans,z;
inline VI F(int*be,int*en){VI a(be,en);sort(a.begin(),a.end());return a;}
map<int,VI>e;
set<int>s;
void dfs(int x){s.insert(x);z|=x==a[n+1];for(int y:e[x])if(!s.count(y))dfs(y);}
int main(){
	ios::sync_with_stdio(0);cin.tie(0);
	cin>>n;for(i=1;i<=n;++i)cin>>a[i],a[n+1]^=a[i];for(i=1;i<=n;++i)cin>>b[i],b[n+1]^=b[i];
	if(F(a+1,a+n+2)!=F(b+1,b+n+2))return puts("-1"),0;
	for(i=1;i<=n;++i)if(a[i]!=b[i])e[a[i]].push_back(b[i]),e[b[i]].push_back(a[i]),++ans;
	for(i=1;i<=n;++i)if(a[i]!=b[i] && !s.count(a[i]))z=0,dfs(a[i]),ans+=!z;
	cout<<ans<<endl;
	return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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>
using namespace std;
typedef vector<int>VI;
const int N=1e5+5;
int a[N],b[N],n,i,ans,z;
inline VI F(int*be,int*en){VI a(be,en);sort(a.begin(),a.end());return a;}
map<int,VI>e;
set<int>s;
void dfs(int x){s.insert(x);z|=x==a[n+1];for(int y:e[x])if(!s.count(y))dfs(y);}
int main(){
	ios::sync_with_stdio(0);cin.tie(0);
	cin>>n;for(i=1;i<=n;++i)cin>>a[i],a[n+1]^=a[i];for(i=1;i<=n;++i)cin>>b[i],b[n+1]^=b[i];
	if(F(a+1,a+n+2)!=F(b+1,b+n+2))return puts("-1"),0;
	for(i=1;i<=n;++i)if(a[i]!=b[i])e[a[i]].push_back(b[i]),e[b[i]].push_back(a[i]),++ans;
	for(i=1;i<=n;++i)if(a[i]!=b[i] && !s.count(a[i]))z=0,dfs(a[i]),ans+=!z;
	cout<<ans<<endl;
	return 0;
}
```