#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N=4e5+10;
int a[N],b[N];
int ta[N],tb[N];
int dp[N],fa[N];
int n,m,ans;
int find(int x){
	return fa[x]==x?x:fa[x]=find(fa[x]);
}
signed main(){
	scanf("%lld",&n);
	for(int i=1;i<=n;++i){
		scanf("%lld",&a[i]);
		a[n+1]^=a[i];
		ta[i]=a[i];
	}
	for(int i=1;i<=n;++i){
		scanf("%lld",&b[i]);
		b[n+1]^=b[i];
		tb[i]=b[i];
	}
	++n;
	ta[n]=a[n];tb[n]=b[n];
	sort(ta+1,ta+n+1);sort(tb+1,tb+n+1);
	for(int i=1;i<n;++i)
		if(ta[i]!=tb[i]){
			printf("-1");
			return 0;
		}
	for(int i=1;i<=n;++i) 
		if(a[i]!=b[i]||i==n){
			dp[++m]=a[i];
			dp[++m]=b[i];
			if(i<n)++ans;
		}
	if(!ans){
		printf("0");
		return 0;
	}
	sort(dp+1,dp+m+1);
	m=unique(dp+1,dp+m+1)-dp-1;
	for(int i=1;i<=m;++i)
		fa[i]=i;
	for(int i=1;i<=n;++i)
		if(a[i]!=b[i]){
			a[i]=lower_bound(dp+1,dp+m+1,a[i])-dp,
			b[i]=lower_bound(dp+1,dp+m+1,b[i])-dp;
			fa[find(a[i])]=find(b[i]);
		}
	for(int i=1;i<=m;++i)
		if(fa[i]==i)
			++ans;
	printf("%lld",ans-1);
	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 int long long
using namespace std;
const int N=4e5+10;
int a[N],b[N];
int ta[N],tb[N];
int dp[N],fa[N];
int n,m,ans;
int find(int x){
	return fa[x]==x?x:fa[x]=find(fa[x]);
}
signed main(){
	scanf("%lld",&n);
	for(int i=1;i<=n;++i){
		scanf("%lld",&a[i]);
		a[n+1]^=a[i];
		ta[i]=a[i];
	}
	for(int i=1;i<=n;++i){
		scanf("%lld",&b[i]);
		b[n+1]^=b[i];
		tb[i]=b[i];
	}
	++n;
	ta[n]=a[n];tb[n]=b[n];
	sort(ta+1,ta+n+1);sort(tb+1,tb+n+1);
	for(int i=1;i<n;++i)
		if(ta[i]!=tb[i]){
			printf("-1");
			return 0;
		}
	for(int i=1;i<=n;++i) 
		if(a[i]!=b[i]||i==n){
			dp[++m]=a[i];
			dp[++m]=b[i];
			if(i<n)++ans;
		}
	if(!ans){
		printf("0");
		return 0;
	}
	sort(dp+1,dp+m+1);
	m=unique(dp+1,dp+m+1)-dp-1;
	for(int i=1;i<=m;++i)
		fa[i]=i;
	for(int i=1;i<=n;++i)
		if(a[i]!=b[i]){
			a[i]=lower_bound(dp+1,dp+m+1,a[i])-dp,
			b[i]=lower_bound(dp+1,dp+m+1,b[i])-dp;
			fa[find(a[i])]=find(b[i]);
		}
	for(int i=1;i<=m;++i)
		if(fa[i]==i)
			++ans;
	printf("%lld",ans-1);
	return 0;
}
```