#include <bits/stdc++.h>
using namespace std;

#pragma GCC optimize("O2,no-stack-protector,unroll-loops,fast-math")

#define F first
#define S second
#define pb push_back
#define SZ(x) (ll)(x.size())
#define all(x) x.begin(),x.end()
#define MP make_pair

typedef int ll;
typedef long double ld;
typedef pair<ll,ll> pll;

//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const ll maxn=1e5+10, maxm=5e3+10, lg=31, mod=1e9+7, inf=1e18;

ll n,a[maxn],b[maxn],ans;
multiset<ll> sa,sb;
map<ll,ll> par;
ll gpar(ll v){return (par[v]==v ? v:par[v]=gpar(par[v]));}
void mrg(ll v,ll u){par[gpar(v)]=gpar(u);}

int main(){
	ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	
	cin>>n;
	for(int i=1;i<=n;i++) cin>>a[i], a[0]^=a[i], sa.insert(a[i]);
	for(int i=1;i<=n;i++) cin>>b[i], b[0]^=b[i], sb.insert(b[i]);
	sa.insert(a[0]), sb.insert(b[0]);
	if(sa!=sb) return cout<<-1,0;
	par[a[0]]=a[0];
	for(int i=1;i<=n;i++)if(a[i]!=b[i]) par[a[i]]=a[i];
	for(int i=1;i<=n;i++)if(a[i]!=b[i]) mrg(a[i],b[i]), ans++;
	for(auto e:par)if(e.F==e.S) ans++;
	cout<<ans-1;

	return 0;
}




### Prompt

Please formulate 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;

#pragma GCC optimize("O2,no-stack-protector,unroll-loops,fast-math")

#define F first
#define S second
#define pb push_back
#define SZ(x) (ll)(x.size())
#define all(x) x.begin(),x.end()
#define MP make_pair

typedef int ll;
typedef long double ld;
typedef pair<ll,ll> pll;

//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const ll maxn=1e5+10, maxm=5e3+10, lg=31, mod=1e9+7, inf=1e18;

ll n,a[maxn],b[maxn],ans;
multiset<ll> sa,sb;
map<ll,ll> par;
ll gpar(ll v){return (par[v]==v ? v:par[v]=gpar(par[v]));}
void mrg(ll v,ll u){par[gpar(v)]=gpar(u);}

int main(){
	ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	
	cin>>n;
	for(int i=1;i<=n;i++) cin>>a[i], a[0]^=a[i], sa.insert(a[i]);
	for(int i=1;i<=n;i++) cin>>b[i], b[0]^=b[i], sb.insert(b[i]);
	sa.insert(a[0]), sb.insert(b[0]);
	if(sa!=sb) return cout<<-1,0;
	par[a[0]]=a[0];
	for(int i=1;i<=n;i++)if(a[i]!=b[i]) par[a[i]]=a[i];
	for(int i=1;i<=n;i++)if(a[i]!=b[i]) mrg(a[i],b[i]), ans++;
	for(auto e:par)if(e.F==e.S) ans++;
	cout<<ans-1;

	return 0;
}




```