#include<bits/stdc++.h>
using namespace std;
int n, a[100007], b[100007], cnt, f[200007], aa[100007], bb[100007], ss[200007], tot;
int read()
{
	int num = 0;
	char c = getchar();
	while (c < '0' || c>'9')c = getchar();
	while (c >= '0' && c <= '9')num = num * 10 + c - '0', c = getchar();
	return num;
}
int find(int x)
{
	if (f[x] == x)return x;
	else return f[x] = find(f[x]);
}
int main()
{
	n = read();
	for (int i = 1; i <= n; i++)
	{
		aa[i] = a[i] = read();
		a[n + 1] ^= a[i];
	}
	for (int i = 1; i <= n; i++)
	{
		bb[i] = b[i] = read();
		b[n + 1] ^= b[i];
	}
	aa[n + 1] = a[n + 1];
	bb[n + 1] = b[n + 1];
	n++;
	sort(aa + 1, aa + n + 1);
	sort(bb + 1, bb + n + 1);
	for (int i = 1; i <= n - 1; i++)
		if (aa[i] != bb[i])
		{
			cout << -1 << endl;
			return 0;
		}
	for(int i=1;i<=n;i++)
		if (a[i] != b[i] || i == n)
		{
			ss[++tot] = a[i];
			ss[++tot] = b[i];
			if (i != n)cnt++;
		}
	sort(ss + 1, ss + tot + 1);
	tot = unique(ss + 1, ss + tot + 1) - ss - 1;
	for (int i = 1; i <= tot; i++)
		f[i] = i;
	for (int i = 1; i <= n; i++)
		if (a[i] != b[i])
		{
			int x = lower_bound(ss + 1, ss + tot + 1, a[i]) - ss;
			int y = lower_bound(ss + 1, ss + tot + 1, b[i]) - ss;
			f[find(x)] = find(y);
		}
	for (int i = 1; i <= n; i++)
		if (f[i] == i)cnt++;
	cout << cnt - 1 << endl;
	return 0;
}
### Prompt

Generate 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;
int n, a[100007], b[100007], cnt, f[200007], aa[100007], bb[100007], ss[200007], tot;
int read()
{
	int num = 0;
	char c = getchar();
	while (c < '0' || c>'9')c = getchar();
	while (c >= '0' && c <= '9')num = num * 10 + c - '0', c = getchar();
	return num;
}
int find(int x)
{
	if (f[x] == x)return x;
	else return f[x] = find(f[x]);
}
int main()
{
	n = read();
	for (int i = 1; i <= n; i++)
	{
		aa[i] = a[i] = read();
		a[n + 1] ^= a[i];
	}
	for (int i = 1; i <= n; i++)
	{
		bb[i] = b[i] = read();
		b[n + 1] ^= b[i];
	}
	aa[n + 1] = a[n + 1];
	bb[n + 1] = b[n + 1];
	n++;
	sort(aa + 1, aa + n + 1);
	sort(bb + 1, bb + n + 1);
	for (int i = 1; i <= n - 1; i++)
		if (aa[i] != bb[i])
		{
			cout << -1 << endl;
			return 0;
		}
	for(int i=1;i<=n;i++)
		if (a[i] != b[i] || i == n)
		{
			ss[++tot] = a[i];
			ss[++tot] = b[i];
			if (i != n)cnt++;
		}
	sort(ss + 1, ss + tot + 1);
	tot = unique(ss + 1, ss + tot + 1) - ss - 1;
	for (int i = 1; i <= tot; i++)
		f[i] = i;
	for (int i = 1; i <= n; i++)
		if (a[i] != b[i])
		{
			int x = lower_bound(ss + 1, ss + tot + 1, a[i]) - ss;
			int y = lower_bound(ss + 1, ss + tot + 1, b[i]) - ss;
			f[find(x)] = find(y);
		}
	for (int i = 1; i <= n; i++)
		if (f[i] == i)cnt++;
	cout << cnt - 1 << endl;
	return 0;
}
```