#include <bits/stdc++.h>
#define il inline
const int N=1e5+5;

int n,m,x,ans,a[N],b[N]; bool p[N];
int to[N+N],nx[N+N],hd[N],sze;
std::unordered_map<int,int> o;

il void add(int u,int v){to[++sze]=v,nx[sze]=hd[u],hd[u]=sze;}

il void dfs(int u){p[u]=1; for (int i=hd[u],v; i; i=nx[i]) if (!p[v=to[i]]) dfs(v);}

int main()
{
	scanf("%d",&n); int i;
	for (i=1; i<=n; i++) scanf("%d",a+i),o[a[i]]++,x^=a[i];
	o[x]++; for (i=1; i<=n; i++) scanf("%d",b+i),o[b[i]]--;
	for (auto &j:o)
		if (j.second<0) return puts("-1"),0;
		else j.second=++m;
	for (i=1; i<=n; i++) if (a[i]!=b[i])
		add(o[a[i]],o[b[i]]),add(o[b[i]],o[a[i]]),ans++;
	
	for (i=1; i<=m; i++) if (!p[i]&&hd[i]) ans++,dfs(i);
	(hd[o[x]])&&(ans--),printf("%d",ans);
	
	return 0;
}
### Prompt

Generate 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>
#define il inline
const int N=1e5+5;

int n,m,x,ans,a[N],b[N]; bool p[N];
int to[N+N],nx[N+N],hd[N],sze;
std::unordered_map<int,int> o;

il void add(int u,int v){to[++sze]=v,nx[sze]=hd[u],hd[u]=sze;}

il void dfs(int u){p[u]=1; for (int i=hd[u],v; i; i=nx[i]) if (!p[v=to[i]]) dfs(v);}

int main()
{
	scanf("%d",&n); int i;
	for (i=1; i<=n; i++) scanf("%d",a+i),o[a[i]]++,x^=a[i];
	o[x]++; for (i=1; i<=n; i++) scanf("%d",b+i),o[b[i]]--;
	for (auto &j:o)
		if (j.second<0) return puts("-1"),0;
		else j.second=++m;
	for (i=1; i<=n; i++) if (a[i]!=b[i])
		add(o[a[i]],o[b[i]]),add(o[b[i]],o[a[i]]),ans++;
	
	for (i=1; i<=m; i++) if (!p[i]&&hd[i]) ans++,dfs(i);
	(hd[o[x]])&&(ans--),printf("%d",ans);
	
	return 0;
}
```