#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm> 
#include <vector>
#include <map>
#define N 100010
using namespace std;
vector<int>Q[N];
map<int,int>d;
bool vis[N];
int n,a[N],b[N],x[N],y[N],cnt,ans;
void dfs(int x){
    vis[x]=1;
    for (int i=0;i<Q[x].size();i++) if (!vis[Q[x][i]]) dfs(Q[x][i]);
}
int main(){
    scanf("%d",&n);

    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]),a[n+1]^=a[i];
        if(!d[a[i]]) d[a[i]]=++cnt;
    }

    if(!d[a[n+1]]) d[a[n+1]]=++cnt;

    for(int i=1;i<=n;i++){
        scanf("%d",&b[i]),b[n+1]^=b[i];
        if(!d[b[i]]) d[b[i]]=++cnt;
    }

    if(!d[b[n+1]]) d[b[n+1]]=++cnt;

    memcpy(x,a,sizeof x),memcpy(y,b,sizeof y);

    sort(x+1,x+n+2),sort(y+1,y+n+2);

    for(int i=1;i<=n;i++) if(x[i]!=y[i]) {printf("-1");return 0;}

    for(int i=1;i<=n+1;i++) a[i]=d[a[i]],b[i]=d[b[i]];

    for(int i=1;i<=n;i++) if(a[i]!=b[i]) ans++,Q[a[i]].push_back(b[i]);

    if(a[n+1]!=b[n+1]) Q[a[n+1]].push_back(b[n+1]);

    for(int i=1;i<cnt;i++) if(Q[i].size()&&!vis[i]) dfs(i),ans++;

    if(!vis[cnt]) ans++;

    printf("%d",ans-1);

	return 0;
}

### 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 <iostream>
#include <cstring>
#include <algorithm> 
#include <vector>
#include <map>
#define N 100010
using namespace std;
vector<int>Q[N];
map<int,int>d;
bool vis[N];
int n,a[N],b[N],x[N],y[N],cnt,ans;
void dfs(int x){
    vis[x]=1;
    for (int i=0;i<Q[x].size();i++) if (!vis[Q[x][i]]) dfs(Q[x][i]);
}
int main(){
    scanf("%d",&n);

    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]),a[n+1]^=a[i];
        if(!d[a[i]]) d[a[i]]=++cnt;
    }

    if(!d[a[n+1]]) d[a[n+1]]=++cnt;

    for(int i=1;i<=n;i++){
        scanf("%d",&b[i]),b[n+1]^=b[i];
        if(!d[b[i]]) d[b[i]]=++cnt;
    }

    if(!d[b[n+1]]) d[b[n+1]]=++cnt;

    memcpy(x,a,sizeof x),memcpy(y,b,sizeof y);

    sort(x+1,x+n+2),sort(y+1,y+n+2);

    for(int i=1;i<=n;i++) if(x[i]!=y[i]) {printf("-1");return 0;}

    for(int i=1;i<=n+1;i++) a[i]=d[a[i]],b[i]=d[b[i]];

    for(int i=1;i<=n;i++) if(a[i]!=b[i]) ans++,Q[a[i]].push_back(b[i]);

    if(a[n+1]!=b[n+1]) Q[a[n+1]].push_back(b[n+1]);

    for(int i=1;i<cnt;i++) if(Q[i].size()&&!vis[i]) dfs(i),ans++;

    if(!vis[cnt]) ans++;

    printf("%d",ans-1);

	return 0;
}

```