#include <bits/stdc++.h>
#define rep(i,n) for ((i)=1;(i)<=(n);(i)++)
#define per(i,n) for ((i)=n;(i)>=(1);(i)--)
using namespace std;
int n,m,i,j,fa[100005],a[100005],b[100005],ta[100005],tb[100005],sa,sb,ans,vis[100005];
vector<int> all;
int fnd(int x){if(fa[x]==x) return x;return fa[x]=fnd(fa[x]);}
int main()
{
	cin>>n;rep(i,n+1) fa[i]=i;
	rep(i,n){cin>>a[i];ta[i]=a[i];sa^=a[i];}
	rep(i,n){cin>>b[i];tb[i]=b[i];sb^=b[i];}
	n++;a[n]=ta[n]=sa;b[n]=tb[n]=sb;
	sort(ta+1,ta+n+1);sort(tb+1,tb+n+1);
	rep(i,n)if(ta[i]!=tb[i]) break;
	if(i<=n){puts("-1");return 0;}
	rep(i,n)if(a[i]!=b[i]) break;
	if(i>n){puts("0");return 0;}
	rep(i,n) all.push_back(a[i]);sort(all.begin(),all.end());
	all.resize(unique(all.begin(),all.end())-all.begin());
	rep(i,n)
	{
		a[i]=upper_bound(all.begin(),all.end(),a[i])-all.begin();
		b[i]=upper_bound(all.begin(),all.end(),b[i])-all.begin();
		if(a[i]!=b[i])
		{
			if(i<n)ans++;
			fa[fnd(a[i])]=fnd(b[i]);
			vis[a[i]]=vis[b[i]]=1;
		}
	}
	vis[a[n]]=vis[b[n]]=1;
	rep(i,n) if(vis[i]&&fa[i]==i) ans++;
	cout<<ans-1<<endl;
	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 rep(i,n) for ((i)=1;(i)<=(n);(i)++)
#define per(i,n) for ((i)=n;(i)>=(1);(i)--)
using namespace std;
int n,m,i,j,fa[100005],a[100005],b[100005],ta[100005],tb[100005],sa,sb,ans,vis[100005];
vector<int> all;
int fnd(int x){if(fa[x]==x) return x;return fa[x]=fnd(fa[x]);}
int main()
{
	cin>>n;rep(i,n+1) fa[i]=i;
	rep(i,n){cin>>a[i];ta[i]=a[i];sa^=a[i];}
	rep(i,n){cin>>b[i];tb[i]=b[i];sb^=b[i];}
	n++;a[n]=ta[n]=sa;b[n]=tb[n]=sb;
	sort(ta+1,ta+n+1);sort(tb+1,tb+n+1);
	rep(i,n)if(ta[i]!=tb[i]) break;
	if(i<=n){puts("-1");return 0;}
	rep(i,n)if(a[i]!=b[i]) break;
	if(i>n){puts("0");return 0;}
	rep(i,n) all.push_back(a[i]);sort(all.begin(),all.end());
	all.resize(unique(all.begin(),all.end())-all.begin());
	rep(i,n)
	{
		a[i]=upper_bound(all.begin(),all.end(),a[i])-all.begin();
		b[i]=upper_bound(all.begin(),all.end(),b[i])-all.begin();
		if(a[i]!=b[i])
		{
			if(i<n)ans++;
			fa[fnd(a[i])]=fnd(b[i]);
			vis[a[i]]=vis[b[i]]=1;
		}
	}
	vis[a[n]]=vis[b[n]]=1;
	rep(i,n) if(vis[i]&&fa[i]==i) ans++;
	cout<<ans-1<<endl;
	return 0;
}
```