#include <bits/stdc++.h>

const int N = 200005;

std::map<int, int> id;

int n, a[N], b[N], c[N], d[N], f[N], ans, tot = 0;
int active(int x) {
	if (id.count(x)) return id[x];
	else return id[x] = ++tot, f[tot] = tot;
}
int find(int x) { return x == f[x] ? x : f[x] = find(f[x]); }
void link(int x, int y) { f[find(x)] = find(y); }

int main() {
	std::ios::sync_with_stdio(0), std::cin.tie(0);
	std::cin >> n;
	for (int i = 1; i <= n; ++i) std::cin >> a[i], a[0] ^= a[i];
	for (int i = 1; i <= n; ++i) std::cin >> b[i], b[0] ^= b[i];
	for (int i = 0; i <= n; ++i) c[i] = a[i], d[i] = b[i];
	std::sort(c, c + n + 1), std::sort(d, d + n + 1);
	for (int i = 0; i <= n; ++i) if (c[i] != d[i]) return std::cout << "-1\n", 0;
	for (int i = 0; i <= n; ++i) if (a[i] != b[i]) link(active(a[i]), active(b[i])), ++ans;
	for (int i = 1; i <= tot; ++i) ans += i == find(i);
	if (a[0] != b[0]) ans -= 2;
	else
		for (int i = 1; i <= n; ++i)
			if (a[i] != b[i] && a[i] == a[0]) {--ans; break;}
	std::cout << ans << '\n';
	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>

const int N = 200005;

std::map<int, int> id;

int n, a[N], b[N], c[N], d[N], f[N], ans, tot = 0;
int active(int x) {
	if (id.count(x)) return id[x];
	else return id[x] = ++tot, f[tot] = tot;
}
int find(int x) { return x == f[x] ? x : f[x] = find(f[x]); }
void link(int x, int y) { f[find(x)] = find(y); }

int main() {
	std::ios::sync_with_stdio(0), std::cin.tie(0);
	std::cin >> n;
	for (int i = 1; i <= n; ++i) std::cin >> a[i], a[0] ^= a[i];
	for (int i = 1; i <= n; ++i) std::cin >> b[i], b[0] ^= b[i];
	for (int i = 0; i <= n; ++i) c[i] = a[i], d[i] = b[i];
	std::sort(c, c + n + 1), std::sort(d, d + n + 1);
	for (int i = 0; i <= n; ++i) if (c[i] != d[i]) return std::cout << "-1\n", 0;
	for (int i = 0; i <= n; ++i) if (a[i] != b[i]) link(active(a[i]), active(b[i])), ++ans;
	for (int i = 1; i <= tot; ++i) ans += i == find(i);
	if (a[0] != b[0]) ans -= 2;
	else
		for (int i = 1; i <= n; ++i)
			if (a[i] != b[i] && a[i] == a[0]) {--ans; break;}
	std::cout << ans << '\n';
	return 0;
}
```