#pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
map<int, int> vis, mp;
int fa[N], a[N], b[N];

int find(int now) { return fa[now] == now ? now : fa[now] = find(fa[now]); }

int main()
{
//	#ifndef ONLINE_JUDGE
//	freopen("duliu.in", "r", stdin);
//	freopen("duliu.out", "w", stdout);
//	#endif
	int n, t = 0;
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) scanf("%d", a + i), vis[a[i]]++, t ^= a[i];
	vis[a[n + 1] = t]++; t = 0;
	for (int i = 1; i <= n; i++)
	{
		scanf("%d", b + i), vis[b[i]]--, t ^= b[i];
		if (vis[b[i]] < 0) return puts("-1"), 0;
	}
	vis[b[n + 1] = t]--; if (vis[t] < 0) return puts("-1"), 0;
	int cnt = 0, lop = 0, all = 0;
	for (int i = 1; i <= n + 1; i++) if (a[i] != b[i])
	{
		all++;
		if (!mp.count(a[i])) mp[a[i]] = ++cnt, fa[cnt] = cnt;
	}
	for (int i = 1; i <= n + 1; i++) if (a[i] != b[i]) fa[find(mp[a[i]])] = find(mp[b[i]]);
	for (int i = 1; i <= cnt; i++) if (fa[i] == i) lop++;
	int ans = all + lop;
	if (a[n + 1] != b[n + 1]) ans -= 2;
	else if (mp.count(a[n + 1])) ans--;
	cout << ans << endl;
//	#ifndef ONLINE_JUDGE
//	fclose(stdin);
//	fclose(stdout);
//	#endif
	return 0;
}
### Prompt

Please formulate 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
#pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
map<int, int> vis, mp;
int fa[N], a[N], b[N];

int find(int now) { return fa[now] == now ? now : fa[now] = find(fa[now]); }

int main()
{
//	#ifndef ONLINE_JUDGE
//	freopen("duliu.in", "r", stdin);
//	freopen("duliu.out", "w", stdout);
//	#endif
	int n, t = 0;
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) scanf("%d", a + i), vis[a[i]]++, t ^= a[i];
	vis[a[n + 1] = t]++; t = 0;
	for (int i = 1; i <= n; i++)
	{
		scanf("%d", b + i), vis[b[i]]--, t ^= b[i];
		if (vis[b[i]] < 0) return puts("-1"), 0;
	}
	vis[b[n + 1] = t]--; if (vis[t] < 0) return puts("-1"), 0;
	int cnt = 0, lop = 0, all = 0;
	for (int i = 1; i <= n + 1; i++) if (a[i] != b[i])
	{
		all++;
		if (!mp.count(a[i])) mp[a[i]] = ++cnt, fa[cnt] = cnt;
	}
	for (int i = 1; i <= n + 1; i++) if (a[i] != b[i]) fa[find(mp[a[i]])] = find(mp[b[i]]);
	for (int i = 1; i <= cnt; i++) if (fa[i] == i) lop++;
	int ans = all + lop;
	if (a[n + 1] != b[n + 1]) ans -= 2;
	else if (mp.count(a[n + 1])) ans--;
	cout << ans << endl;
//	#ifndef ONLINE_JUDGE
//	fclose(stdin);
//	fclose(stdout);
//	#endif
	return 0;
}
```