#include <bits/stdc++.h>
#define eb emplace_back
#define sz(V) ((int)(V).size())
#define allv(V) ((V).begin()),((V).end())
#define sorv(V) sort(allv(V))
#define univ(V) (V).erase(unique(allv(V)),(V).end())
using namespace std;

const int MAXN = 100055;

struct DJF {
	DJF() { init(); }
	int ud[MAXN*2];

	void init() { iota(ud, ud+MAXN*2, 0); }
	int uf(int i) { return i == ud[i] ? i : (ud[i] = uf(ud[i])); }
	void uf(int a, int b) { ud[uf(b)] = uf(a); }
} djf;

vector<int> XV, UV;

int A[MAXN], B[MAXN];

int N, Ans;

int getI(int x) { return int(lower_bound(allv(XV), x) - XV.begin()); }

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0);

	cin >> N;
	for(int i = 1; i <= N; i++) {
		cin >> A[i];
		A[0] ^= A[i];
	}
	for(int i = 1; i <= N; i++) {
		cin >> B[i];
		B[0] ^= B[i];
	}
	{
		vector<int> AV, BV;
		for(int i = 0; i <= N; i++) AV.eb(A[i]);
		for(int i = 0; i <= N; i++) BV.eb(B[i]);
		sorv(AV); sorv(BV);
		if(AV != BV) {
			puts("-1");
			exit(0);
		}
	}
	XV.eb(A[0]); XV.eb(B[0]);
	for(int i = 1; i <= N; i++) if(A[i] != B[i]) {
		XV.eb(A[i]);
		XV.eb(B[i]);
		Ans++;
	}
	sorv(XV); univ(XV);
	for(int i = 1; i <= N; i++) if(A[i] != B[i])
		djf.uf(getI(A[i]), getI(B[i]));
	for(int i = sz(XV); i--;) UV.eb(djf.uf(i));
	sorv(UV); univ(UV);

	cout << Ans + sz(UV) - 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>
#define eb emplace_back
#define sz(V) ((int)(V).size())
#define allv(V) ((V).begin()),((V).end())
#define sorv(V) sort(allv(V))
#define univ(V) (V).erase(unique(allv(V)),(V).end())
using namespace std;

const int MAXN = 100055;

struct DJF {
	DJF() { init(); }
	int ud[MAXN*2];

	void init() { iota(ud, ud+MAXN*2, 0); }
	int uf(int i) { return i == ud[i] ? i : (ud[i] = uf(ud[i])); }
	void uf(int a, int b) { ud[uf(b)] = uf(a); }
} djf;

vector<int> XV, UV;

int A[MAXN], B[MAXN];

int N, Ans;

int getI(int x) { return int(lower_bound(allv(XV), x) - XV.begin()); }

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0);

	cin >> N;
	for(int i = 1; i <= N; i++) {
		cin >> A[i];
		A[0] ^= A[i];
	}
	for(int i = 1; i <= N; i++) {
		cin >> B[i];
		B[0] ^= B[i];
	}
	{
		vector<int> AV, BV;
		for(int i = 0; i <= N; i++) AV.eb(A[i]);
		for(int i = 0; i <= N; i++) BV.eb(B[i]);
		sorv(AV); sorv(BV);
		if(AV != BV) {
			puts("-1");
			exit(0);
		}
	}
	XV.eb(A[0]); XV.eb(B[0]);
	for(int i = 1; i <= N; i++) if(A[i] != B[i]) {
		XV.eb(A[i]);
		XV.eb(B[i]);
		Ans++;
	}
	sorv(XV); univ(XV);
	for(int i = 1; i <= N; i++) if(A[i] != B[i])
		djf.uf(getI(A[i]), getI(B[i]));
	for(int i = sz(XV); i--;) UV.eb(djf.uf(i));
	sorv(UV); univ(UV);

	cout << Ans + sz(UV) - 1 << endl;
	return 0;
}
```