#include<cstdio>
#include<algorithm>
#include<cstring>
#include<map>
#define MN 200005
using namespace std;
int n,a[MN],b[MN],ans,num,head[MN],tt,sum[MN];
struct edge{int to,next;}g[MN];
map<int,int>mp;bool vis[MN];
void ins(int u,int v){g[++num].next=head[u];head[u]=num;g[num].to=v;}
void dfs(int u){vis[u]=1;for(int i=head[u];i;i=g[i].next)if(!vis[g[i].to])dfs(g[i].to);}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",&a[i]),a[n+1]^=a[i];
	for(int i=1;i<=n;i++)scanf("%d",&b[i]),b[n+1]^=b[i];
	for(int i=1;i<=n+1;i++)mp[a[i]]++,mp[b[i]]--;
	for(int i=1;i<=n+1;i++)if(mp[a[i]]){puts("-1");return 0;}
	for(int i=1;i<=n+1;i++)if(!mp[a[i]])mp[a[i]]=++tt;
	for(int i=1;i<=n+1;i++)if(a[i]!=b[i])ins(mp[a[i]],mp[b[i]]),ans++,++sum[mp[a[i]]];
	for(int i=1;i<=tt;i++)if(!vis[i]&&sum[i])dfs(i),ans++;
	ans-=(sum[mp[a[n+1]]]!=0)+(a[n+1]!=b[n+1]);
	printf("%d\n",ans);
	return 0;
}
### Prompt

Your task is to create 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<cstdio>
#include<algorithm>
#include<cstring>
#include<map>
#define MN 200005
using namespace std;
int n,a[MN],b[MN],ans,num,head[MN],tt,sum[MN];
struct edge{int to,next;}g[MN];
map<int,int>mp;bool vis[MN];
void ins(int u,int v){g[++num].next=head[u];head[u]=num;g[num].to=v;}
void dfs(int u){vis[u]=1;for(int i=head[u];i;i=g[i].next)if(!vis[g[i].to])dfs(g[i].to);}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",&a[i]),a[n+1]^=a[i];
	for(int i=1;i<=n;i++)scanf("%d",&b[i]),b[n+1]^=b[i];
	for(int i=1;i<=n+1;i++)mp[a[i]]++,mp[b[i]]--;
	for(int i=1;i<=n+1;i++)if(mp[a[i]]){puts("-1");return 0;}
	for(int i=1;i<=n+1;i++)if(!mp[a[i]])mp[a[i]]=++tt;
	for(int i=1;i<=n+1;i++)if(a[i]!=b[i])ins(mp[a[i]],mp[b[i]]),ans++,++sum[mp[a[i]]];
	for(int i=1;i<=tt;i++)if(!vis[i]&&sum[i])dfs(i),ans++;
	ans-=(sum[mp[a[n+1]]]!=0)+(a[n+1]!=b[n+1]);
	printf("%d\n",ans);
	return 0;
}
```