#include<bits/stdc++.h>
using namespace std;
#define MAXN	100005
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
map <int, int> cnt, num;
vector <int> g[MAXN];
int ans, s, n, tot;
int a[MAXN], b[MAXN];
bool visited[MAXN];
void dfs(int x) {
	visited[x] = true;
	ans += g[x].size();
	for (unsigned i = 0; i < g[x].size(); i++)
		if (!visited[g[x][i]]) dfs(g[x][i]);
}
int main() {
	read(n); s = 0;
	for (int i = 1; i <= n; i++) {
		read(a[i]);
		s ^= a[i];
		cnt[a[i]]++;
	}
	cnt[s]++;
	for (int i = 1; i <= n; i++) {
		read(b[i]);
		if (--cnt[b[i]] < 0) {
			printf("-1");
			return 0;
		}
	}
	num[s] = tot = 1;
	for (int i = 1; i <= n; i++)
		if (num[a[i]] == 0) num[a[i]] = ++tot;
	for (int i = 1; i <= n; i++)
		if (num[b[i]] != num[a[i]]) g[num[b[i]]].push_back(num[a[i]]);
	dfs(num[s]);
	for (int i = 1; i <= n; i++) {
		int tmp = num[a[i]];
		if (!visited[tmp] && g[tmp].size()) {
			dfs(tmp);
			ans++;
		}
	}
	cout << ans << endl;
	return 0;
}
### Prompt

Your challenge is to write 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;
#define MAXN	100005
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
map <int, int> cnt, num;
vector <int> g[MAXN];
int ans, s, n, tot;
int a[MAXN], b[MAXN];
bool visited[MAXN];
void dfs(int x) {
	visited[x] = true;
	ans += g[x].size();
	for (unsigned i = 0; i < g[x].size(); i++)
		if (!visited[g[x][i]]) dfs(g[x][i]);
}
int main() {
	read(n); s = 0;
	for (int i = 1; i <= n; i++) {
		read(a[i]);
		s ^= a[i];
		cnt[a[i]]++;
	}
	cnt[s]++;
	for (int i = 1; i <= n; i++) {
		read(b[i]);
		if (--cnt[b[i]] < 0) {
			printf("-1");
			return 0;
		}
	}
	num[s] = tot = 1;
	for (int i = 1; i <= n; i++)
		if (num[a[i]] == 0) num[a[i]] = ++tot;
	for (int i = 1; i <= n; i++)
		if (num[b[i]] != num[a[i]]) g[num[b[i]]].push_back(num[a[i]]);
	dfs(num[s]);
	for (int i = 1; i <= n; i++) {
		int tmp = num[a[i]];
		if (!visited[tmp] && g[tmp].size()) {
			dfs(tmp);
			ans++;
		}
	}
	cout << ans << endl;
	return 0;
}
```