#include <cstdio>
#include <vector>
#include <map>

const int MN = 100005;

int N, A[MN], B[MN], X, cnt, M, Ans;
std::map<int, int> mp;

int vis[MN];
std::vector<int> G[MN];
void DFS(int u) {
	vis[u] = 1;
	for (int v : G[u]) if (!vis[v]) DFS(v);
}

int main() {
	scanf("%d", &N);
	for (int i = 1; i <= N; ++i) scanf("%d", &A[i]), ++mp[A[i]], X ^= A[i];
	for (int i = 1; i <= N; ++i) scanf("%d", &B[i]), --mp[B[i]];
	++mp[X];
	for (auto p : mp) if (p.second < 0) return puts("-1"), 0;
	for (auto &p : mp) p.second = ++cnt;
	for (int i = 1; i <= N; ++i) if (A[i] != B[i]) {
		int u = mp[B[i]], v = mp[A[i]];
		G[u].push_back(v);
		G[v].push_back(u);
		++M;
	}
	for (int i = 1; i <= cnt; ++i) if (!G[i].empty())
		if (!vis[i]) ++Ans, DFS(i);
	if (!G[mp[X]].empty()) --Ans;
	printf("%d\n", M + Ans);
	return 0;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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 <vector>
#include <map>

const int MN = 100005;

int N, A[MN], B[MN], X, cnt, M, Ans;
std::map<int, int> mp;

int vis[MN];
std::vector<int> G[MN];
void DFS(int u) {
	vis[u] = 1;
	for (int v : G[u]) if (!vis[v]) DFS(v);
}

int main() {
	scanf("%d", &N);
	for (int i = 1; i <= N; ++i) scanf("%d", &A[i]), ++mp[A[i]], X ^= A[i];
	for (int i = 1; i <= N; ++i) scanf("%d", &B[i]), --mp[B[i]];
	++mp[X];
	for (auto p : mp) if (p.second < 0) return puts("-1"), 0;
	for (auto &p : mp) p.second = ++cnt;
	for (int i = 1; i <= N; ++i) if (A[i] != B[i]) {
		int u = mp[B[i]], v = mp[A[i]];
		G[u].push_back(v);
		G[v].push_back(u);
		++M;
	}
	for (int i = 1; i <= cnt; ++i) if (!G[i].empty())
		if (!vis[i]) ++Ans, DFS(i);
	if (!G[mp[X]].empty()) --Ans;
	printf("%d\n", M + Ans);
	return 0;
}
```