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;

int a[100009],b[100009],c[100009],fa[100009],va[100009]={0},cva[100009]={0},cvb[100009]={0},xore=0;
bool is[100009]={0},vis=0;

int gf(int u){
	if(fa[u]==u){return u;}
	va[fa[u]]+=va[u];va[u]=0;
	return fa[u]=gf(fa[u]);
}

int main(){
#ifndef ONLINE_JUDGE
	freopen("duliu.in","r",stdin);
	freopen("duliu.out","w",stdout);
#endif
	int n,res=0,tot,apos,bpos;
	scanf("%d",&n);
	for(int i=1;i<=n;i=-(~i)){scanf("%d",&a[i]);c[i]=a[i];xore^=a[i];}	
	sort(c+1,c+1+n);tot=unique(c+1,c+1+n)-c-1;
	for(int i=1;i<=tot;i=-(~i)){fa[i]=i;}fa[tot+1]=tot+1;
	for(int i=1;i<=n;i=-(~i)){
		scanf("%d",&b[i]);
		if(a[i]==b[i]){continue;}
		apos=lower_bound(c+1,c+1+tot,a[i])-c;
		bpos=lower_bound(c+1,c+1+tot,b[i])-c;
		if(c[bpos]!=b[i]){
			if(vis||b[i]!=xore){printf("-1\n");return 0;}
			else{vis=true;res--;bpos=tot+1;}//本质 xore 不参与离散化，也不参与并查集统计
		}
		cva[apos]++;cvb[bpos]++;
//		fprintf(stdout,"%d[%d:%d %d[%d:%d %d %d\n",a[i],c[apos],apos,b[i],c[bpos],bpos,tot,xore);
		if(gf(apos)!=gf(bpos)){va[fa[apos]]+=va[fa[bpos]];va[fa[bpos]]=0;fa[fa[bpos]]=fa[apos];}//注意顺序
		va[fa[bpos]]++;
	}
	for(int i=1;i<=tot;i=-(~i)){
		gf(i);
		if(c[i]==xore){vis=true;res--;cvb[i]--;}
		//不能直接用vis判断
		if(cva[i]<cvb[i]){printf("-1\n");return 0;}
	}
	for(int i=1;i<=tot;i=-(~i)){
		if(va[i]){res+=va[i]+1;}//fprintf(stdout,"%d:%d\n",c[i],va[i]+1);}
	}
	printf("%d\n",((res<0)?0:res));
	return 0;
}