#include<cstdio> 
#include<algorithm>
#include<map>
#define fo(i,a,b) for(i=a;i<=b;i++)
using namespace std;
const int maxn=100000+10;
map<int,int> f;
int a[maxn],b[maxn],c[maxn],d[maxn],fa[maxn*2];
int i,j,k,l,t,n,m,tot,ans;
int getfa(int x){
    return fa[x]==x?x:fa[x]=getfa(fa[x]);
}
int main(){
    scanf("%d",&n);
    fo(i,1,n) scanf("%d",&a[i]);
    fo(i,1,n) scanf("%d",&b[i]);
    t=0;
    fo(i,1,n) t^=a[i];
    a[n+1]=t;
    t=0;
    fo(i,1,n) t^=b[i];
    b[++n]=t;
    fo(i,1,n) c[i]=a[i],d[i]=b[i];
    sort(c+1,c+n+1);
    sort(d+1,d+n+1);
    fo(i,1,n+1)
        if (i>n||c[i]!=d[i]) break;
    if (i<=n){
        printf("-1\n");
        return 0;
    }
    fo(i,1,n)
        if (a[i]!=b[i]||i==n){
            if (i<n) ans++;
            if (!f[a[i]]) f[a[i]]=++tot;
            if (!f[b[i]]) f[b[i]]=++tot;
        }
    if (!ans){
        printf("0\n");
        return 0;
    }
    fo(i,1,tot) fa[i]=i;
    fo(i,1,n)
        if (a[i]!=b[i]) fa[getfa(f[a[i]])]=getfa(f[b[i]]);
    fo(i,1,tot)
        if (fa[i]==i) ans++;
    printf("%d\n",ans-1);
}
### Prompt

Develop a solution in CPP to the problem described below:

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<algorithm>
#include<map>
#define fo(i,a,b) for(i=a;i<=b;i++)
using namespace std;
const int maxn=100000+10;
map<int,int> f;
int a[maxn],b[maxn],c[maxn],d[maxn],fa[maxn*2];
int i,j,k,l,t,n,m,tot,ans;
int getfa(int x){
    return fa[x]==x?x:fa[x]=getfa(fa[x]);
}
int main(){
    scanf("%d",&n);
    fo(i,1,n) scanf("%d",&a[i]);
    fo(i,1,n) scanf("%d",&b[i]);
    t=0;
    fo(i,1,n) t^=a[i];
    a[n+1]=t;
    t=0;
    fo(i,1,n) t^=b[i];
    b[++n]=t;
    fo(i,1,n) c[i]=a[i],d[i]=b[i];
    sort(c+1,c+n+1);
    sort(d+1,d+n+1);
    fo(i,1,n+1)
        if (i>n||c[i]!=d[i]) break;
    if (i<=n){
        printf("-1\n");
        return 0;
    }
    fo(i,1,n)
        if (a[i]!=b[i]||i==n){
            if (i<n) ans++;
            if (!f[a[i]]) f[a[i]]=++tot;
            if (!f[b[i]]) f[b[i]]=++tot;
        }
    if (!ans){
        printf("0\n");
        return 0;
    }
    fo(i,1,tot) fa[i]=i;
    fo(i,1,n)
        if (a[i]!=b[i]) fa[getfa(f[a[i]])]=getfa(f[b[i]]);
    fo(i,1,tot)
        if (fa[i]==i) ans++;
    printf("%d\n",ans-1);
}
```