#include <cstdio>
#include <algorithm>
#include <iostream>
#include <vector>
#define MN 101000

int a[MN], b[MN];
int c[MN], d[MN];
int s = 0, s2 = 0;
bool vis[MN];
std::vector<int> v[MN];
bool ok = 1;
int ans = 0, now = 0;

void dfs(int x)
{
	if(vis[x]) return; 
	vis[x] = 1;
	now += v[x].size();
	if(c[x] == s) ok = 0;
	for(auto &y : v[x]) dfs(y);
}

int main()
{
	int n; scanf("%d", &n);
	for(int i = 1; i <= n; i++) scanf("%d", &a[i]), s ^= a[i], c[i] = a[i];
	for(int i = 1; i <= n; i++) scanf("%d", &b[i]), s2 ^= b[i], d[i] = b[i];
	c[n + 1] = s; d[n + 1] = s2;
	std::sort(c + 1, c + n + 2);
	std::sort(d + 1, d + n + 2);
	for(int i = 1; i <= n + 1; i++) if(c[i] != d[i]) return 0 * puts("-1");
	int N = std::unique(c + 1, c + n + 2) - c - 1;
	for(int i = 1; i <= n; i++)
	{
		int A = std::lower_bound(c + 1, c + N + 1, a[i]) - c;
		int B = std::lower_bound(c + 1, c + N + 1, b[i]) - c;
		if(A != B) v[A].push_back(B), v[B].push_back(A);
	}
	for(int i = 1; i <= N; i++)
		if(!vis[i]) 
		{
			now = 0; ok = 1;
			dfs(i);
			if(now) ans += now / 2 + ok;
		}
	printf("%d\n", ans);
}

### Prompt

Please create a solution in Cpp 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 <cstdio>
#include <algorithm>
#include <iostream>
#include <vector>
#define MN 101000

int a[MN], b[MN];
int c[MN], d[MN];
int s = 0, s2 = 0;
bool vis[MN];
std::vector<int> v[MN];
bool ok = 1;
int ans = 0, now = 0;

void dfs(int x)
{
	if(vis[x]) return; 
	vis[x] = 1;
	now += v[x].size();
	if(c[x] == s) ok = 0;
	for(auto &y : v[x]) dfs(y);
}

int main()
{
	int n; scanf("%d", &n);
	for(int i = 1; i <= n; i++) scanf("%d", &a[i]), s ^= a[i], c[i] = a[i];
	for(int i = 1; i <= n; i++) scanf("%d", &b[i]), s2 ^= b[i], d[i] = b[i];
	c[n + 1] = s; d[n + 1] = s2;
	std::sort(c + 1, c + n + 2);
	std::sort(d + 1, d + n + 2);
	for(int i = 1; i <= n + 1; i++) if(c[i] != d[i]) return 0 * puts("-1");
	int N = std::unique(c + 1, c + n + 2) - c - 1;
	for(int i = 1; i <= n; i++)
	{
		int A = std::lower_bound(c + 1, c + N + 1, a[i]) - c;
		int B = std::lower_bound(c + 1, c + N + 1, b[i]) - c;
		if(A != B) v[A].push_back(B), v[B].push_back(A);
	}
	for(int i = 1; i <= N; i++)
		if(!vis[i]) 
		{
			now = 0; ok = 1;
			dfs(i);
			if(now) ans += now / 2 + ok;
		}
	printf("%d\n", ans);
}

```