#include<bits/stdc++.h>
#define L long long
#define pb push_back
using namespace std;
int n,a[100010],b[100010],x[200010],p,q;
map<int,int> h;
vector<int> w[200010],A;
bool u[200010];
inline int dfs(int i)
{
	int j,k=0;
	u[i]=1;
	for(j=0;j<w[i].size();j++)
	  {
	   if(!u[w[i][j]])
	     k+=dfs(w[i][j]);
	   if(w[i][j]!=i)
	     k++;
	  }
	return k;
}
int main()
{
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	int i,j,k=0,l;
	scanf("%d",&n);
	for(i=1;i<=n;i++)
	  scanf("%d",&a[i]);
	for(i=1;i<=n;i++)
	  scanf("%d",&b[i]);
	for(i=1;i<=n;i++)
	  {
	   k^=a[i];
	   if(!h[a[i]])
	     h[a[i]]=++p;
	   a[i]=h[a[i]];
	   if(!h[b[i]])
	     h[b[i]]=++p;
	   b[i]=h[b[i]];
	   w[a[i]].pb(b[i]);
	   x[a[i]]++;
	   x[b[i]]--;
	  }
	if(!h[k])
	  h[k]=++p;
	k=h[k];
	for(i=1,j=0;i<=p;i++)
	  if(x[i]>0)
	    j+=x[i];
	if(j>1 || j==1 && x[k]!=-1)
	  {
	   printf("-1\n");
	   return 0;
	  }
	q=-1;
	if(j==1)
	  {
	   for(i=1;i<=p;i++)
	     if(x[i]==1)
	       j=i;
	   w[k].pb(j);
	   q--;
	  }
	for(i=1;i<=p;i++)
	  if(!u[i])
	    {
	     l=dfs(i);
	     if(l)
	       q+=l+1;
	     else if(i==k)
	       q++;
		}
	printf("%d\n",q);
	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 L long long
#define pb push_back
using namespace std;
int n,a[100010],b[100010],x[200010],p,q;
map<int,int> h;
vector<int> w[200010],A;
bool u[200010];
inline int dfs(int i)
{
	int j,k=0;
	u[i]=1;
	for(j=0;j<w[i].size();j++)
	  {
	   if(!u[w[i][j]])
	     k+=dfs(w[i][j]);
	   if(w[i][j]!=i)
	     k++;
	  }
	return k;
}
int main()
{
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	int i,j,k=0,l;
	scanf("%d",&n);
	for(i=1;i<=n;i++)
	  scanf("%d",&a[i]);
	for(i=1;i<=n;i++)
	  scanf("%d",&b[i]);
	for(i=1;i<=n;i++)
	  {
	   k^=a[i];
	   if(!h[a[i]])
	     h[a[i]]=++p;
	   a[i]=h[a[i]];
	   if(!h[b[i]])
	     h[b[i]]=++p;
	   b[i]=h[b[i]];
	   w[a[i]].pb(b[i]);
	   x[a[i]]++;
	   x[b[i]]--;
	  }
	if(!h[k])
	  h[k]=++p;
	k=h[k];
	for(i=1,j=0;i<=p;i++)
	  if(x[i]>0)
	    j+=x[i];
	if(j>1 || j==1 && x[k]!=-1)
	  {
	   printf("-1\n");
	   return 0;
	  }
	q=-1;
	if(j==1)
	  {
	   for(i=1;i<=p;i++)
	     if(x[i]==1)
	       j=i;
	   w[k].pb(j);
	   q--;
	  }
	for(i=1;i<=p;i++)
	  if(!u[i])
	    {
	     l=dfs(i);
	     if(l)
	       q+=l+1;
	     else if(i==k)
	       q++;
		}
	printf("%d\n",q);
	return 0;
}

```