#include<bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
  int n;
  cin>>n;
  int a[n+1],b[n+1];
  for(int i=0;i<n;i++) cin>>a[i];
  for(int i=0;i<n;i++) cin>>b[i];
  a[n]=b[n]=0;
  for(int i=0;i<n;i++)
    a[n]^=a[i],b[n]^=b[i];
  map<int,int> x,y;
  for(int i=0;i<=n;i++) x[a[i]]++;
  for(int i=0;i<=n;i++) y[b[i]]++;
  if(x!=y){
    cout<<-1<<endl;
    return 0;
  }
  
  vector<int> v(n+1);
  for(int i=0;i<=n;i++) v[i]=a[i];
  sort(v.begin(),v.end());
  v.erase(unique(v.begin(),v.end()),v.end());
  map<int,int> m;
  for(int i=0;i<(int)v.size();i++) m[v[i]]=i;
  vector<int> G[v.size()];
  int used[v.size()];
  memset(used,0,sizeof(used));
  for(int i=0;i<=n;i++)
    if(a[i]!=b[i]) G[m[a[i]]].push_back(m[b[i]]);
  int cnt=0;
  for(int i=0;i<(int)v.size();i++){
    if(used[i]) continue;
    if(G[i].empty()) continue;
    cnt++;
    queue<int> q;
    q.push(i);
    while(!q.empty()){
      int k=q.front();q.pop();
      if(used[k]) continue;
      used[k]=1;
      for(int u:G[k])
	if(!used[u]) q.push(u);
    }
  }
  if(G[m[a[n]]].empty()) cnt++;
  if(a[n]!=b[n]){
    if(G[m[b[n]]].empty()) cnt++;
    else cnt--;
  }
  int ans=0;
  for(int i=0;i<(int)v.size();i++) ans+=G[i].size();
  cout<<ans+cnt-1<<endl;
  return 0;
}

### Prompt

Your challenge is to write 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;
#define int long long
signed main(){
  int n;
  cin>>n;
  int a[n+1],b[n+1];
  for(int i=0;i<n;i++) cin>>a[i];
  for(int i=0;i<n;i++) cin>>b[i];
  a[n]=b[n]=0;
  for(int i=0;i<n;i++)
    a[n]^=a[i],b[n]^=b[i];
  map<int,int> x,y;
  for(int i=0;i<=n;i++) x[a[i]]++;
  for(int i=0;i<=n;i++) y[b[i]]++;
  if(x!=y){
    cout<<-1<<endl;
    return 0;
  }
  
  vector<int> v(n+1);
  for(int i=0;i<=n;i++) v[i]=a[i];
  sort(v.begin(),v.end());
  v.erase(unique(v.begin(),v.end()),v.end());
  map<int,int> m;
  for(int i=0;i<(int)v.size();i++) m[v[i]]=i;
  vector<int> G[v.size()];
  int used[v.size()];
  memset(used,0,sizeof(used));
  for(int i=0;i<=n;i++)
    if(a[i]!=b[i]) G[m[a[i]]].push_back(m[b[i]]);
  int cnt=0;
  for(int i=0;i<(int)v.size();i++){
    if(used[i]) continue;
    if(G[i].empty()) continue;
    cnt++;
    queue<int> q;
    q.push(i);
    while(!q.empty()){
      int k=q.front();q.pop();
      if(used[k]) continue;
      used[k]=1;
      for(int u:G[k])
	if(!used[u]) q.push(u);
    }
  }
  if(G[m[a[n]]].empty()) cnt++;
  if(a[n]!=b[n]){
    if(G[m[b[n]]].empty()) cnt++;
    else cnt--;
  }
  int ans=0;
  for(int i=0;i<(int)v.size();i++) ans+=G[i].size();
  cout<<ans+cnt-1<<endl;
  return 0;
}

```