#include<bits/stdc++.h>
#define rep(i,a,b) for (int i=(a); i<=(b); i++)
#define per(i,a,b) for (int i=(a); i>=(b); i--)
using namespace std;

const int maxn = 100005;
int a[maxn], b[maxn], pa[maxn], sz[maxn];
int n, ans, flag = 1, cnt;
multiset<int> s;
map<int,int> M;

inline int getpa(int x) {
	return pa[x] == x ? x : pa[x] = getpa(pa[x]);
}

int main() {
	scanf("%d", &n);
	rep (i, 1, n) {
		scanf("%d", &a[i]);
		a[n+1] ^= a[i];
	}
	rep (i, 1, n) {
		scanf("%d", &b[i]);
		b[n+1] ^= b[i];
	}
	n++;
	rep (i, 1, n) s.insert(a[i]);
	rep (i, 1, n) {
		if (s.find(b[i]) == s.end()) 
			return puts("-1"), 0;
		s.erase(s.find(b[i]));
	}
	rep (i, 1, n) if (i == n || a[i] != b[i]) {
		if (!M[a[i]]) a[i] = M[a[i]] = ++cnt;
		else a[i] = M[a[i]];
		if (!M[b[i]]) b[i] = M[b[i]] = ++cnt;
		else b[i] = M[b[i]];
		if (i < n) ans++;
	}
	else a[i] = b[i] = 0;
	rep (i, 1, cnt) pa[i] = i;
	rep (i, 1, n) if (a[i])
		pa[getpa(a[i])] = getpa(b[i]);
	rep (i, 1, cnt) if (pa[i] == i) ans++;
	printf("%d\n", ans - 1);
	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>
#define rep(i,a,b) for (int i=(a); i<=(b); i++)
#define per(i,a,b) for (int i=(a); i>=(b); i--)
using namespace std;

const int maxn = 100005;
int a[maxn], b[maxn], pa[maxn], sz[maxn];
int n, ans, flag = 1, cnt;
multiset<int> s;
map<int,int> M;

inline int getpa(int x) {
	return pa[x] == x ? x : pa[x] = getpa(pa[x]);
}

int main() {
	scanf("%d", &n);
	rep (i, 1, n) {
		scanf("%d", &a[i]);
		a[n+1] ^= a[i];
	}
	rep (i, 1, n) {
		scanf("%d", &b[i]);
		b[n+1] ^= b[i];
	}
	n++;
	rep (i, 1, n) s.insert(a[i]);
	rep (i, 1, n) {
		if (s.find(b[i]) == s.end()) 
			return puts("-1"), 0;
		s.erase(s.find(b[i]));
	}
	rep (i, 1, n) if (i == n || a[i] != b[i]) {
		if (!M[a[i]]) a[i] = M[a[i]] = ++cnt;
		else a[i] = M[a[i]];
		if (!M[b[i]]) b[i] = M[b[i]] = ++cnt;
		else b[i] = M[b[i]];
		if (i < n) ans++;
	}
	else a[i] = b[i] = 0;
	rep (i, 1, cnt) pa[i] = i;
	rep (i, 1, n) if (a[i])
		pa[getpa(a[i])] = getpa(b[i]);
	rep (i, 1, cnt) if (pa[i] == i) ans++;
	printf("%d\n", ans - 1);
	return 0;
}
```