#include <bits/stdc++.h>

#define For(i, j, k) for(int i = j; i <= k; i++)

using namespace std;

const int N = 1e5 + 10;

int A[N], B[N], n;
int fa[N];

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

void Union(int u, int v){
	u = find(u), v = find(v);
	if(u ^ v) fa[u] = v;
}

map<int, int> M;
map<int, queue<int> > T;

int main(){
	
	scanf("%d", &n);

	int S = 0;
	For(i, 1, n) scanf("%d", &A[i]), S ^= A[i];
	For(i, 1, n) scanf("%d", &B[i]);
	
	int x = 0;
	For(i, 1, n) if(A[i] != B[i]) A[++x] = A[i], B[x] = B[i];
	n = x;

	For(i, 1, n) fa[i] = i, M[A[i]]++;

	int pos = 0;
	For(i, 1, n) if(!M[B[i]]){
		if(pos || B[i] != S){
			puts("-1");
			return 0;
		}
		pos = i;
	}
	else --M[B[i]];

	int ans = n;
	For(i, 1, n) if(B[i] == S){
		--ans;
		break;
	}

	For(i, 1, n){
		auto &q = T[A[i]];
		if(!q.empty()) Union(i, q.front());
		q.push(i);
	}
	For(i, 1, n){
		auto &q = T[B[i]];
		if(!q.empty()) Union(i, q.front()), q.pop();
	}

	For(i, 1, n) if(fa[i] == i) ++ans;
	printf("%d\n", ans);

	return 0;
}
### 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 <bits/stdc++.h>

#define For(i, j, k) for(int i = j; i <= k; i++)

using namespace std;

const int N = 1e5 + 10;

int A[N], B[N], n;
int fa[N];

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

void Union(int u, int v){
	u = find(u), v = find(v);
	if(u ^ v) fa[u] = v;
}

map<int, int> M;
map<int, queue<int> > T;

int main(){
	
	scanf("%d", &n);

	int S = 0;
	For(i, 1, n) scanf("%d", &A[i]), S ^= A[i];
	For(i, 1, n) scanf("%d", &B[i]);
	
	int x = 0;
	For(i, 1, n) if(A[i] != B[i]) A[++x] = A[i], B[x] = B[i];
	n = x;

	For(i, 1, n) fa[i] = i, M[A[i]]++;

	int pos = 0;
	For(i, 1, n) if(!M[B[i]]){
		if(pos || B[i] != S){
			puts("-1");
			return 0;
		}
		pos = i;
	}
	else --M[B[i]];

	int ans = n;
	For(i, 1, n) if(B[i] == S){
		--ans;
		break;
	}

	For(i, 1, n){
		auto &q = T[A[i]];
		if(!q.empty()) Union(i, q.front());
		q.push(i);
	}
	For(i, 1, n){
		auto &q = T[B[i]];
		if(!q.empty()) Union(i, q.front()), q.pop();
	}

	For(i, 1, n) if(fa[i] == i) ++ans;
	printf("%d\n", ans);

	return 0;
}
```