#include <bits/stdc++.h>
using namespace std;
#define x first
#define y second
#define dbg(x) cout << #x << '=' << x << '\n';
#define ll long long
#define pi pair<int,int>
#define pl pair<long long,long long>
#define lg length()
#define pb push_back
ifstream in("file.in");
ofstream out("file.out");
#define MAXN 100005
#define INF 1000000005
#define LINF 1000000000000000005

int n,a[100005],b[100005],t,ans,x[100005],y[100005];

map <int,vector<int>> m;
map <int,int> v;

void DFS(int nod){
    if(v[nod]) return;
    v[nod]=1;
    for(int i : m[nod]){
        DFS(i);
    }
}

int32_t main(){
    ios_base :: sync_with_stdio(0);
    cin >> n;
    for(int i=1;i<=n;i++){
        cin >> a[i];
        a[n+1]^=a[i];
        x[i]=a[i];
    }
    for(int i=1;i<=n;i++){
        cin >> b[i];
        b[n+1]^=b[i];
        y[i]=b[i];
    }
    x[n+1]=a[n+1]; y[n+1]=b[n+1];
    n++;
    sort(x+1,x+n+1);
    sort(y+1,y+n+1);
    for(int i=1;i<=n;i++){
        if(x[i]!=y[i]){
            cout << -1;
            return 0;
        }
    }

    ans=-2;

    for(int i=1;i<=n;i++){
        m[a[i]].push_back(b[i]);
        m[b[i]].push_back(a[i]);
        if(a[i]!=b[i] || i==n) ans++;
    }

    for(int i=1;i<=n;i++){
        if((a[i]!=b[i] || i==n) && !v[a[i]]) DFS(a[i]),ans++;
    }

    cout << ans;
}

### Prompt

Please provide a Cpp coded solution 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 <bits/stdc++.h>
using namespace std;
#define x first
#define y second
#define dbg(x) cout << #x << '=' << x << '\n';
#define ll long long
#define pi pair<int,int>
#define pl pair<long long,long long>
#define lg length()
#define pb push_back
ifstream in("file.in");
ofstream out("file.out");
#define MAXN 100005
#define INF 1000000005
#define LINF 1000000000000000005

int n,a[100005],b[100005],t,ans,x[100005],y[100005];

map <int,vector<int>> m;
map <int,int> v;

void DFS(int nod){
    if(v[nod]) return;
    v[nod]=1;
    for(int i : m[nod]){
        DFS(i);
    }
}

int32_t main(){
    ios_base :: sync_with_stdio(0);
    cin >> n;
    for(int i=1;i<=n;i++){
        cin >> a[i];
        a[n+1]^=a[i];
        x[i]=a[i];
    }
    for(int i=1;i<=n;i++){
        cin >> b[i];
        b[n+1]^=b[i];
        y[i]=b[i];
    }
    x[n+1]=a[n+1]; y[n+1]=b[n+1];
    n++;
    sort(x+1,x+n+1);
    sort(y+1,y+n+1);
    for(int i=1;i<=n;i++){
        if(x[i]!=y[i]){
            cout << -1;
            return 0;
        }
    }

    ans=-2;

    for(int i=1;i<=n;i++){
        m[a[i]].push_back(b[i]);
        m[b[i]].push_back(a[i]);
        if(a[i]!=b[i] || i==n) ans++;
    }

    for(int i=1;i<=n;i++){
        if((a[i]!=b[i] || i==n) && !v[a[i]]) DFS(a[i]),ans++;
    }

    cout << ans;
}

```