#include<iostream>
#include<cstdio>
#include<map>
#include<vector>
#define fi first
#define se second
#define N 100100
using namespace std;

int n,a[N],b[N],cnt,ans,ta,tb,now,fa[N];
map<int,int>mm,ca,cb;
map<int,int>::iterator ia,ib;

int ff(int u){return u==fa[u]?u:fa[u]=ff(fa[u]);}
inline void in(int u){if(!mm[u]) mm[u]=++cnt;}

int main()
{
	int i,j,p,q;
	cin>>n;
	for(i=1;i<=n;i++) scanf("%d",&a[i]),ta^=a[i],fa[i]=i;ca[ta]++;
	for(i=1;i<=n;i++)
	{
		scanf("%d",&b[i]);
		if(b[i]!=a[i])
		{
			in(a[i]),in(b[i]);
			ca[a[i]]++;
			cb[b[i]]++;
			ans++;
		}
		tb^=b[i];
	}
	cb[tb]++;
	in(ta),in(tb);
	for(ia=ca.begin(),ib=cb.begin();ia!=ca.end();ia++,ib++)
	{
		if((*ia).fi!=(*ib).fi||(*ia).se!=(*ib).se)
		{
			puts("-1");
			return 0;
		}
	}
	for(i=1;i<=n;i++)
	{
		if(a[i]!=b[i])
		{
			fa[ff(mm[a[i]])]=ff(mm[b[i]]);
		}
	}
	if(ta!=tb) fa[ff(mm[ta])]=ff(mm[tb]);
	for(i=1;i<=cnt;i++) if(i==ff(i)) ans++;
	cout<<ans-1;
}
### 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<iostream>
#include<cstdio>
#include<map>
#include<vector>
#define fi first
#define se second
#define N 100100
using namespace std;

int n,a[N],b[N],cnt,ans,ta,tb,now,fa[N];
map<int,int>mm,ca,cb;
map<int,int>::iterator ia,ib;

int ff(int u){return u==fa[u]?u:fa[u]=ff(fa[u]);}
inline void in(int u){if(!mm[u]) mm[u]=++cnt;}

int main()
{
	int i,j,p,q;
	cin>>n;
	for(i=1;i<=n;i++) scanf("%d",&a[i]),ta^=a[i],fa[i]=i;ca[ta]++;
	for(i=1;i<=n;i++)
	{
		scanf("%d",&b[i]);
		if(b[i]!=a[i])
		{
			in(a[i]),in(b[i]);
			ca[a[i]]++;
			cb[b[i]]++;
			ans++;
		}
		tb^=b[i];
	}
	cb[tb]++;
	in(ta),in(tb);
	for(ia=ca.begin(),ib=cb.begin();ia!=ca.end();ia++,ib++)
	{
		if((*ia).fi!=(*ib).fi||(*ia).se!=(*ib).se)
		{
			puts("-1");
			return 0;
		}
	}
	for(i=1;i<=n;i++)
	{
		if(a[i]!=b[i])
		{
			fa[ff(mm[a[i]])]=ff(mm[b[i]]);
		}
	}
	if(ta!=tb) fa[ff(mm[ta])]=ff(mm[tb]);
	for(i=1;i<=cnt;i++) if(i==ff(i)) ans++;
	cout<<ans-1;
}
```