#include <bits/stdc++.h>
using namespace std;
 
#define endl '\n'
#define die(x) return cout << x << endl, 0
#define all(v) v.begin(), v.end()
#define sz(x) (int)(x.size())
const int N = 1e5 + 5;

int A[N], B[N], P[N];

int GetPar(int v) { return P[v] == v ? v : P[v] = GetPar(P[v]); }
void Merge(int v, int u) {
	v = GetPar(v), u = GetPar(u);
	if (u == v) return;
	P[v] = u;
}

int main() {
 
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);

	int n, x = 0, ans = 0; cin >> n;
	vector<int> a, b;
	for (int i = 0; i < n; i++) cin >> A[i], x ^= A[i];
	A[n] = x,  x = 0;
	for (int i = 0; i < n; i++) cin >> B[i], x ^= B[i];
	B[n] = x, x = 0;
	for (int i = 0; i < n; i++) if (A[i] != B[i]) a.push_back(A[i]), b.push_back(B[i]), ans++;
	a.push_back(A[n]), b.push_back(B[n]);
	sort(all(a)), sort(all(b));
	if (a != b) die(-1);
	a.resize(unique(all(a)) - a.begin()), b.resize(unique(all(b)) - b.begin());
	for (int i = 0; i < sz(a); i++) P[i] = i;
	for (int i = 0; i <= n; i++) if (A[i] != B[i]) {
		int v = lower_bound(all(a), A[i]) - a.begin();
		int u = lower_bound(all(b), B[i]) - b.begin();
		Merge(v, u);
	}
	for (int i = 0; i < sz(a); i++) ans += (GetPar(i) == i);
	cout << ans - 1 << endl;

	return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 endl '\n'
#define die(x) return cout << x << endl, 0
#define all(v) v.begin(), v.end()
#define sz(x) (int)(x.size())
const int N = 1e5 + 5;

int A[N], B[N], P[N];

int GetPar(int v) { return P[v] == v ? v : P[v] = GetPar(P[v]); }
void Merge(int v, int u) {
	v = GetPar(v), u = GetPar(u);
	if (u == v) return;
	P[v] = u;
}

int main() {
 
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);

	int n, x = 0, ans = 0; cin >> n;
	vector<int> a, b;
	for (int i = 0; i < n; i++) cin >> A[i], x ^= A[i];
	A[n] = x,  x = 0;
	for (int i = 0; i < n; i++) cin >> B[i], x ^= B[i];
	B[n] = x, x = 0;
	for (int i = 0; i < n; i++) if (A[i] != B[i]) a.push_back(A[i]), b.push_back(B[i]), ans++;
	a.push_back(A[n]), b.push_back(B[n]);
	sort(all(a)), sort(all(b));
	if (a != b) die(-1);
	a.resize(unique(all(a)) - a.begin()), b.resize(unique(all(b)) - b.begin());
	for (int i = 0; i < sz(a); i++) P[i] = i;
	for (int i = 0; i <= n; i++) if (A[i] != B[i]) {
		int v = lower_bound(all(a), A[i]) - a.begin();
		int u = lower_bound(all(b), B[i]) - b.begin();
		Merge(v, u);
	}
	for (int i = 0; i < sz(a); i++) ans += (GetPar(i) == i);
	cout << ans - 1 << endl;

	return 0;
}

```