#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<map>
#include<vector>
using namespace std;

const int N=200005;

int n,sz,ans;
int a[N],b[N],c[N];
bool vis[N];
map<int,int> c1,c2;
vector<int> lev[N];

void dfs(int k){
	int j;
	vis[k]=1;
	for (j=0;j<lev[k].size();j++)
		if (!vis[lev[k][j]]) dfs(lev[k][j]);
}

int main(){
	int i;
	scanf("%d",&n);
	for (i=1;i<=n;i++) scanf("%d",&a[i]),a[n+1]^=a[i];
	for (i=1;i<=n;i++) scanf("%d",&b[i]),b[n+1]^=b[i];
	n++;
	for (i=1;i<=n;i++) c1[a[i]]++;
	for (i=1;i<=n;i++) c2[b[i]]++;
	for (i=1;i<=n;i++) if (c1[a[i]]!=c2[a[i]]){
		printf("-1\n");
		return 0;
	}
	for (i=1;i<=n;i++) c[++sz]=a[i];
	for (i=1;i<=n;i++) c[++sz]=b[i];
	sort(c+1,c+1+sz); sz=unique(c+1,c+1+sz)-c-1;
	for (i=1;i<=n;i++) a[i]=lower_bound(c+1,c+1+sz,a[i])-c;
	for (i=1;i<=n;i++) b[i]=lower_bound(c+1,c+1+sz,b[i])-c;
	for (i=1;i<n;i++) if (a[i]!=b[i]){
		lev[a[i]].push_back(b[i]);
		ans++;
	}
	lev[a[n]].push_back(b[n]);
	if (!ans){
		printf("0\n");
		return 0;
	}
	for (i=1;i<=n;i++)
		if (!vis[a[i]]&&lev[a[i]].size()){
			dfs(a[i]);
			ans++;
		}
	printf("%d\n",ans-1);
	return 0;
}
### Prompt

Your challenge is to write 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<cmath>
#include<algorithm>
#include<cstdlib>
#include<map>
#include<vector>
using namespace std;

const int N=200005;

int n,sz,ans;
int a[N],b[N],c[N];
bool vis[N];
map<int,int> c1,c2;
vector<int> lev[N];

void dfs(int k){
	int j;
	vis[k]=1;
	for (j=0;j<lev[k].size();j++)
		if (!vis[lev[k][j]]) dfs(lev[k][j]);
}

int main(){
	int i;
	scanf("%d",&n);
	for (i=1;i<=n;i++) scanf("%d",&a[i]),a[n+1]^=a[i];
	for (i=1;i<=n;i++) scanf("%d",&b[i]),b[n+1]^=b[i];
	n++;
	for (i=1;i<=n;i++) c1[a[i]]++;
	for (i=1;i<=n;i++) c2[b[i]]++;
	for (i=1;i<=n;i++) if (c1[a[i]]!=c2[a[i]]){
		printf("-1\n");
		return 0;
	}
	for (i=1;i<=n;i++) c[++sz]=a[i];
	for (i=1;i<=n;i++) c[++sz]=b[i];
	sort(c+1,c+1+sz); sz=unique(c+1,c+1+sz)-c-1;
	for (i=1;i<=n;i++) a[i]=lower_bound(c+1,c+1+sz,a[i])-c;
	for (i=1;i<=n;i++) b[i]=lower_bound(c+1,c+1+sz,b[i])-c;
	for (i=1;i<n;i++) if (a[i]!=b[i]){
		lev[a[i]].push_back(b[i]);
		ans++;
	}
	lev[a[n]].push_back(b[n]);
	if (!ans){
		printf("0\n");
		return 0;
	}
	for (i=1;i<=n;i++)
		if (!vis[a[i]]&&lev[a[i]].size()){
			dfs(a[i]);
			ans++;
		}
	printf("%d\n",ans-1);
	return 0;
}
```