#include <bits/stdc++.h>
#define NO  puts("-1"),exit(0)
using namespace std;

const int N=100010;
int n,m,a[N],b[N],c[2*N],ufs[2*N],ans;
map <int,int> M;

int find(int x){return ufs[x]==x?x:ufs[x]=find(ufs[x]);}

void work()
{
	scanf("%d",&n);
	for (int i=1; i<=n; i++)  scanf("%d",&a[i]),a[0]^=a[i];
	for (int i=1; i<=n; i++)  scanf("%d",&b[i]),b[0]^=b[i];
	for (int i=0; i<=n; i++)
		if (a[i]!=b[i])
			M[a[i]]++,M[b[i]]--,c[++m]=a[i],c[++m]=b[i];
	for (map <int,int> ::iterator x=M.begin(); x!=M.end(); x++)
		if ((*x).second)  NO;
	sort(c+1,c+m+1),m=unique(c+1,c+m+1)-c-1;
	for (int i=1; i<=n; i++)  a[i]=lower_bound(c+1,c+m+1,a[i])-c;
	for (int i=1; i<=n; i++)  b[i]=lower_bound(c+1,c+m+1,b[i])-c;
	ans=m;
	for (int i=1; i<=m; i++)  ufs[i]=i;
	for (int i=1; i<=n; i++)
		if (a[i]!=b[i])
			{
				if (find(a[i])!=find(b[i]))  ufs[find(a[i])]=find(b[i]);
				else  ans++;
			}
	for (int i=1; i<=m; i++)  if (c[i]==a[0])  ans--;
	printf("%d",ans);
}

int main()
{
	work();
	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 NO  puts("-1"),exit(0)
using namespace std;

const int N=100010;
int n,m,a[N],b[N],c[2*N],ufs[2*N],ans;
map <int,int> M;

int find(int x){return ufs[x]==x?x:ufs[x]=find(ufs[x]);}

void work()
{
	scanf("%d",&n);
	for (int i=1; i<=n; i++)  scanf("%d",&a[i]),a[0]^=a[i];
	for (int i=1; i<=n; i++)  scanf("%d",&b[i]),b[0]^=b[i];
	for (int i=0; i<=n; i++)
		if (a[i]!=b[i])
			M[a[i]]++,M[b[i]]--,c[++m]=a[i],c[++m]=b[i];
	for (map <int,int> ::iterator x=M.begin(); x!=M.end(); x++)
		if ((*x).second)  NO;
	sort(c+1,c+m+1),m=unique(c+1,c+m+1)-c-1;
	for (int i=1; i<=n; i++)  a[i]=lower_bound(c+1,c+m+1,a[i])-c;
	for (int i=1; i<=n; i++)  b[i]=lower_bound(c+1,c+m+1,b[i])-c;
	ans=m;
	for (int i=1; i<=m; i++)  ufs[i]=i;
	for (int i=1; i<=n; i++)
		if (a[i]!=b[i])
			{
				if (find(a[i])!=find(b[i]))  ufs[find(a[i])]=find(b[i]);
				else  ans++;
			}
	for (int i=1; i<=m; i++)  if (c[i]==a[0])  ans--;
	printf("%d",ans);
}

int main()
{
	work();
	return 0;
}

```