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<bits/stdc++.h>
#define read() Read<int>()
#define lowbit(i) i&(-i)
using namespace std;
typedef long long ll;
const int N=1e5+5;
namespace pb_ds{   
    namespace io{
        const int MaxBuff=1<<15;
        const int Output=1<<23;
        char B[MaxBuff],*S=B,*T=B;
		#define getc() ((S==T)&&(T=(S=B)+fread(B,1,MaxBuff,stdin),S==T)?0:*S++)
        char Out[Output],*iter=Out;
        inline void flush(){
            fwrite(Out,1,iter-Out,stdout);
            iter=Out;
        }
    }
    template<class Type> inline Type Read(){
        using namespace io;
        register char ch;
        register Type ans=0; 
        register bool neg=0;
        while(ch=getc(),(ch<'0' || ch>'9') && ch!='-');
        ch=='-'?neg=1:ans=ch-'0';
        while(ch=getc(),'0'<= ch && ch<='9') ans=ans*10+ch-'0';
        return neg?-ans:ans;
    }
    template<class Type> inline void Print(register Type x,register char ch='\n'){
        using namespace io;
        if(!x) *iter++='0';
        else{
            if(x<0) *iter++='-',x=-x;
            static int s[100]; 
            register int t=0;
            while(x) s[++t]=x%10,x/=10;
            while(t) *iter++='0'+s[t--];
        }
        *iter++=ch;
    }
}
using namespace pb_ds;
int n,a[N],b[N],ans,sta,k=-1;
map<int,int>cnt;
int main(){
//	freopen("duliu.in","r",stdin);
//	freopen("duliu.out","w",stdout);
	n=read();
	for (int i=1;i<=n;++i) a[i]=read(),++cnt[a[i]],sta^=a[i];
	++cnt[sta];
	for (int i=1;i<=n;++i) b[i]=read(),--cnt[b[i]];
	for (int i=1;i<=n;++i){
		ans+=(a[i]!=b[i]);
		if (cnt[b[i]]<0){
			printf("-1\n");
			return 0;
		}	
	}
	if (!ans){
		printf("0\n");
		return 0;	
	}
	++ans;
	for (int i=1;i<=n;++i)
		if (b[i]==sta && b[i]!=a[i]){
			--ans;
			break;
		}
	printf("%d\n",ans);
	return 0;
}