#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#define N 200010
using namespace std;
int n, a[N], b[N], l, s1[N], s2[N], f[N], x1, y1, ans, flag[N];
map<int, int>mp;
inline int gether(int x){return f[x]?f[x]=gether(f[x]):x;}
int main(){
	scanf("%d", &n);
	for(int i=1; i<=n; i++)scanf("%d", &a[i]);
	a[n+1]=0; for(int i=1; i<=n; i++)a[n+1]^=a[i];
	for(int i=1; i<=n; i++)scanf("%d", &b[i]);
	b[n+1]=0; for(int i=1; i<=n; i++)b[n+1]^=b[i];
	n++; l=0; mp.clear();
	for(int i=1; i<=n; i++)if(!mp[a[i]])mp[a[i]]=++l;
	for(int i=1; i<=n; i++)if(!mp[b[i]])mp[b[i]]=++l;
	memset(s1, 0, sizeof(s1)); memset(s2, 0, sizeof(s2));
	for(int i=1; i<=n; i++){s1[mp[a[i]]]++; s2[mp[b[i]]]++;}
	for(int i=1; i<=l; i++)if(s1[i]!=s2[i]){printf("-1"); return 0;}
	ans=0; if(a[n]!=b[n])ans--;
	memset(f, 0, sizeof(f)); memset(flag, 0, sizeof(flag));
	for(int i=1; i<=n; i++)if(a[i]!=b[i]){
		flag[mp[a[i]]]=flag[mp[b[i]]]=1;
		ans++; x1=gether(mp[a[i]]); y1=gether(mp[b[i]]);
		if(x1!=y1)f[y1]=x1;
	}
	for(int i=1; i<=l; i++)if(flag[i]&&!f[i])ans++;
	if(flag[mp[a[n]]])ans--;
	printf("%d", ans);
	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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<cstring>
#include<algorithm>
#include<map>
#define N 200010
using namespace std;
int n, a[N], b[N], l, s1[N], s2[N], f[N], x1, y1, ans, flag[N];
map<int, int>mp;
inline int gether(int x){return f[x]?f[x]=gether(f[x]):x;}
int main(){
	scanf("%d", &n);
	for(int i=1; i<=n; i++)scanf("%d", &a[i]);
	a[n+1]=0; for(int i=1; i<=n; i++)a[n+1]^=a[i];
	for(int i=1; i<=n; i++)scanf("%d", &b[i]);
	b[n+1]=0; for(int i=1; i<=n; i++)b[n+1]^=b[i];
	n++; l=0; mp.clear();
	for(int i=1; i<=n; i++)if(!mp[a[i]])mp[a[i]]=++l;
	for(int i=1; i<=n; i++)if(!mp[b[i]])mp[b[i]]=++l;
	memset(s1, 0, sizeof(s1)); memset(s2, 0, sizeof(s2));
	for(int i=1; i<=n; i++){s1[mp[a[i]]]++; s2[mp[b[i]]]++;}
	for(int i=1; i<=l; i++)if(s1[i]!=s2[i]){printf("-1"); return 0;}
	ans=0; if(a[n]!=b[n])ans--;
	memset(f, 0, sizeof(f)); memset(flag, 0, sizeof(flag));
	for(int i=1; i<=n; i++)if(a[i]!=b[i]){
		flag[mp[a[i]]]=flag[mp[b[i]]]=1;
		ans++; x1=gether(mp[a[i]]); y1=gether(mp[b[i]]);
		if(x1!=y1)f[y1]=x1;
	}
	for(int i=1; i<=l; i++)if(flag[i]&&!f[i])ans++;
	if(flag[mp[a[n]]])ans--;
	printf("%d", ans);
	return 0;
}
```