#include<bits/stdc++.h>
using namespace std;
const int N=2e5+10;
int gi() {
    int x=0,o=1;char ch=getchar();
    while((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
    if(ch=='-') o=-1,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*o;
}
int n,a[N],b[N],c[N],d[N],fa[N],st[N],tp=0;
int getf(int x) {
    return x==fa[x]?x:fa[x]=getf(fa[x]);
}
int main() {
    n=gi();
    int sum=0;
    for(int i=1;i<=n;i++) a[i]=gi(),sum^=a[i];
    a[n+1]=sum;sum=0;
    for(int i=1;i<=n;i++) b[i]=gi(),sum^=b[i];
    b[n+1]=sum;
    ++n;
    for(int i=1;i<=n;i++) c[i]=a[i],d[i]=b[i];
    sort(c+1,c+n+1),sort(d+1,d+n+1);
    for(int i=1;i<=n;i++) if(c[i]!=d[i]) return puts("-1"),0;
    int ans=0;
    for(int i=1;i<=n;i++)
        if(a[i]!=b[i]||i==n) {
            st[++tp]=a[i],st[++tp]=b[i];
            if(i<n) ++ans;
        }
    sort(st+1,st+tp+1);
    tp=unique(st+1,st+tp+1)-st-1;
    for(int i=1;i<=tp;i++) fa[i]=i;
    for(int i=1;i<n;i++) {
        a[i]=lower_bound(st+1,st+tp+1,a[i])-st;
        b[i]=lower_bound(st+1,st+tp+1,b[i])-st;
        fa[getf(a[i])]=getf(b[i]);
    }
    for(int i=1;i<=tp;i++) if(getf(i)==i) ++ans;
    printf("%d\n",ans-1);
    return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

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<bits/stdc++.h>
using namespace std;
const int N=2e5+10;
int gi() {
    int x=0,o=1;char ch=getchar();
    while((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
    if(ch=='-') o=-1,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*o;
}
int n,a[N],b[N],c[N],d[N],fa[N],st[N],tp=0;
int getf(int x) {
    return x==fa[x]?x:fa[x]=getf(fa[x]);
}
int main() {
    n=gi();
    int sum=0;
    for(int i=1;i<=n;i++) a[i]=gi(),sum^=a[i];
    a[n+1]=sum;sum=0;
    for(int i=1;i<=n;i++) b[i]=gi(),sum^=b[i];
    b[n+1]=sum;
    ++n;
    for(int i=1;i<=n;i++) c[i]=a[i],d[i]=b[i];
    sort(c+1,c+n+1),sort(d+1,d+n+1);
    for(int i=1;i<=n;i++) if(c[i]!=d[i]) return puts("-1"),0;
    int ans=0;
    for(int i=1;i<=n;i++)
        if(a[i]!=b[i]||i==n) {
            st[++tp]=a[i],st[++tp]=b[i];
            if(i<n) ++ans;
        }
    sort(st+1,st+tp+1);
    tp=unique(st+1,st+tp+1)-st-1;
    for(int i=1;i<=tp;i++) fa[i]=i;
    for(int i=1;i<n;i++) {
        a[i]=lower_bound(st+1,st+tp+1,a[i])-st;
        b[i]=lower_bound(st+1,st+tp+1,b[i])-st;
        fa[getf(a[i])]=getf(b[i]);
    }
    for(int i=1;i<=tp;i++) if(getf(i)==i) ++ans;
    printf("%d\n",ans-1);
    return 0;
}

```