#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>

namespace SaoTi
{
	const int N=101000;
	bool cmp(int *s,int *t,int n)
	{
		static int A[N],B[N];
		memcpy(A+1,s+1,n*4),std::sort(A+1,A+n+1);
		memcpy(B+1,t+1,n*4),std::sort(B+1,B+n+1);
		for(int i=1;i<=n;i++)if(A[i]!=B[i])return 0;
		return 1;
	}
	std::map<int,int> id;
	int fa[N],sel[N];
	void reset(int n){for(int i=1;i<=n;i++)fa[i]=i;}
	int ask(int p){return p==fa[p]?p:fa[p]=ask(fa[p]);}
	void link(int u,int v){fa[ask(u)]=ask(v);}

	int A[N],B[N];
	int n,tot;
	void solve()
	{
		scanf("%d",&n);
		A[n+1]=0;
		for(int i=1;i<=n;i++)
			scanf("%d",A+i),A[n+1]^=A[i];
		B[n+1]=0;
		for(int i=1;i<=n;i++)
			scanf("%d",B+i),B[n+1]^=B[i];
		if(!cmp(A,B,n+1)){printf("-1\n");return ;}

		int ans=0;

		reset(n+1);
		for(int i=1,u,v;i<=n;i++)
			if(A[i]!=B[i])
			{
				ans++;
				if(!id.count(A[i]))id[A[i]]=++tot;
				if(!id.count(B[i]))id[B[i]]=++tot;
				u=id[A[i]],v=id[B[i]];
				link(u,v);
			}

		for(int i=1;i<=tot;i++)
			if(fa[i]==i)ans++;
		if(!id.count(A[n+1]))ans++;
		ans--;
		printf("%d\n",ans);
	}
}

int main()
{
	SaoTi::solve();
	return 0;
}

### Prompt

Generate a cpp solution 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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>

namespace SaoTi
{
	const int N=101000;
	bool cmp(int *s,int *t,int n)
	{
		static int A[N],B[N];
		memcpy(A+1,s+1,n*4),std::sort(A+1,A+n+1);
		memcpy(B+1,t+1,n*4),std::sort(B+1,B+n+1);
		for(int i=1;i<=n;i++)if(A[i]!=B[i])return 0;
		return 1;
	}
	std::map<int,int> id;
	int fa[N],sel[N];
	void reset(int n){for(int i=1;i<=n;i++)fa[i]=i;}
	int ask(int p){return p==fa[p]?p:fa[p]=ask(fa[p]);}
	void link(int u,int v){fa[ask(u)]=ask(v);}

	int A[N],B[N];
	int n,tot;
	void solve()
	{
		scanf("%d",&n);
		A[n+1]=0;
		for(int i=1;i<=n;i++)
			scanf("%d",A+i),A[n+1]^=A[i];
		B[n+1]=0;
		for(int i=1;i<=n;i++)
			scanf("%d",B+i),B[n+1]^=B[i];
		if(!cmp(A,B,n+1)){printf("-1\n");return ;}

		int ans=0;

		reset(n+1);
		for(int i=1,u,v;i<=n;i++)
			if(A[i]!=B[i])
			{
				ans++;
				if(!id.count(A[i]))id[A[i]]=++tot;
				if(!id.count(B[i]))id[B[i]]=++tot;
				u=id[A[i]],v=id[B[i]];
				link(u,v);
			}

		for(int i=1;i<=tot;i++)
			if(fa[i]==i)ans++;
		if(!id.count(A[n+1]))ans++;
		ans--;
		printf("%d\n",ans);
	}
}

int main()
{
	SaoTi::solve();
	return 0;
}

```