#include<bits/stdc++.h>
#define pb push_back
using namespace std;
const int N=1e5+50;
int n,a[N],b[N],c[N],c2[N],m,ans;bool vis[N];
vector<int>v[N];
void dfs(int x){
	vis[x]=1;
	for(int i=0,y;i<v[x].size();i++)
		if(!vis[y=v[x][i]])dfs(y);
}
int main(){
	scanf("%d",&n);n++;
	for(int i=1;i<n;i++)scanf("%d",&a[i]),a[n]^=a[i];
	for(int i=1;i<n;i++)scanf("%d",&b[i]),b[n]^=b[i];
	for(int i=1;i<=n;i++)c[i]=a[i],c2[i]=b[i];
	sort(c+1,c+n+1);sort(c2+1,c2+n+1);
	for(int i=1;i<=n;i++)if(c[i]!=c2[i])puts("-1"),exit(0);
	m=unique(c+1,c+n+1)-c-1;
	for(int i=1;i<=n;i++)a[i]=lower_bound(c+1,c+m+1,a[i])-c,b[i]=lower_bound(c+1,c+m+1,b[i])-c;
	for(int i=1;i<=n;i++)if(a[i]!=b[i])v[a[i]].pb(b[i]),ans++;
	for(int i=1;i<=m;i++)if(!vis[i]&&v[i].size())dfs(i),ans++;ans-=2;
	if(a[n]==b[n]){
		if(!v[a[n]].size())ans+=2;
		else ans++;
	}
	cout<<ans;
	return 0;
}
### Prompt

Please create a solution in cpp 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 pb push_back
using namespace std;
const int N=1e5+50;
int n,a[N],b[N],c[N],c2[N],m,ans;bool vis[N];
vector<int>v[N];
void dfs(int x){
	vis[x]=1;
	for(int i=0,y;i<v[x].size();i++)
		if(!vis[y=v[x][i]])dfs(y);
}
int main(){
	scanf("%d",&n);n++;
	for(int i=1;i<n;i++)scanf("%d",&a[i]),a[n]^=a[i];
	for(int i=1;i<n;i++)scanf("%d",&b[i]),b[n]^=b[i];
	for(int i=1;i<=n;i++)c[i]=a[i],c2[i]=b[i];
	sort(c+1,c+n+1);sort(c2+1,c2+n+1);
	for(int i=1;i<=n;i++)if(c[i]!=c2[i])puts("-1"),exit(0);
	m=unique(c+1,c+n+1)-c-1;
	for(int i=1;i<=n;i++)a[i]=lower_bound(c+1,c+m+1,a[i])-c,b[i]=lower_bound(c+1,c+m+1,b[i])-c;
	for(int i=1;i<=n;i++)if(a[i]!=b[i])v[a[i]].pb(b[i]),ans++;
	for(int i=1;i<=m;i++)if(!vis[i]&&v[i].size())dfs(i),ans++;ans-=2;
	if(a[n]==b[n]){
		if(!v[a[n]].size())ans+=2;
		else ans++;
	}
	cout<<ans;
	return 0;
}
```