Problem:
Score : 1000 points

Problem Statement

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.

Sample Input 1

3
0 1 2
3 1 0

Sample Output 1

2

At first, the XOR of all the elements of a is 3.
If we replace a_1 with 3, a becomes (3, 1, 2).

Now, the XOR of all the elements of a is 0.
If we replace a_3 with 0, a becomes (3, 1, 0), which matches b.

Sample Input 2

3
0 1 2
0 1 2

Sample Output 2

0

Sample Input 3

2
1 1
0 0

Sample Output 3

-1

Sample Input 4

4
0 1 2 3
1 0 3 2

Sample Output 4

5

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<cstdio>
#include<algorithm>
using namespace std;
const int N=100010;
int n,ck=2,suma,sumb,tota,totb,fxor,ans,cnt,f[N],a[N],b[N],lsha[N],lshb[N];
int find(int x){
	return x==f[x]?x:f[x]=find(f[x]);
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d",a+i);
		lsha[i]=a[i];
		suma^=a[i];
	}
	for(int i=1;i<=n;i++){
		scanf("%d",b+i);
		lshb[i]=b[i];
		sumb^=b[i];
		if(a[i]!=b[i])
			ans++;
	}
	if(!ans){
		puts("0");
		return 0;
	}
	++n;
	lsha[n]=a[n]=suma,lshb[n]=b[n]=sumb;
	sort(lsha+1,lsha+n+1);
	sort(lshb+1,lshb+n+1);
	for(int i=1;i<=n;i++){
		if(lsha[i]!=lshb[i]){
			puts("-1");
			return 0;
		}
	}
	tota=unique(lsha+1,lsha+n+1)-(lsha+1);
	for(int i=1;i<=tota;i++)
		f[i]=i;
	for(int i=1;i<=n;i++){
		int x=lower_bound(lsha+1,lsha+tota+1,a[i])-lsha,y=lower_bound(lsha+1,lsha+tota+1,b[i])-lsha;
		int fx=find(x),fy=find(y);
		if(fx!=fy)
			f[fx]=fy;
	}
	fxor=find(lower_bound(lsha+1,lsha+n+1,suma)-lsha);
	for(int i=1;i<=n;i++){
		int fi=find(i);
		if(i==fi&&a[i]!=b[i])
			cnt++;
		if(fi==fxor)
			ck--;
	}
	printf("%d",ans+max(ck,0)+max(cnt-1,0));
	return 0;
}