#include<bits/stdc++.h>
const int MAXN = 2e5 + 10;
using namespace std;
int n;
int ans=0,sum=0;
int A[MAXN], B[MAXN], ta[MAXN], tb[MAXN], sa, sb, tot = 0, Do[MAXN], fa[MAXN];
int find(int x) {
    return fa[x] == x ? fa[x] : fa[x] = find(fa[x]);
}
int unionn(int x, int y) {
    fa[x] = y;
}
int main() {
    scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%d",&A[i]),sa^=A[i];A[n+1]=sa;
    for(int i=1;i<=n;i++)scanf("%d",&B[i]),sb^=B[i];B[n+1]=sb;
    memcpy(ta,A,sizeof(A));memcpy(tb,B,sizeof(B));
    n++;sort(ta+1,ta+n+1);sort(tb+1,tb+n+1);
    for(int i=1;i<n;i++)if(ta[i]!=tb[i])return puts("-1"), 0;
    for(int i=1;i<=n;i++)
        if(A[i]!=B[i]||(i==n)){
            Do[++sum]=A[i]; 
			Do[++sum]=B[i];
            if(i<n)ans++;
        }
    if(ans==0)return puts("0"), 0;
    sort(Do+1,Do+sum+1);
    sum=unique(Do+1,Do+sum+1)-Do-1;
    for(int i=1;i<=sum;i++)fa[i]=i;
    for(int i=1;i<=n;i++)
        if(A[i]!=B[i]) {
            A[i]=lower_bound(Do+1,Do+sum+1,A[i])-Do;
            B[i]=lower_bound(Do+1,Do+sum+1,B[i])-Do;
            unionn(find(A[i]),find(B[i]));
        }
    for(int i=1;i<=sum;i++)ans+=(fa[i]==i);
    printf("%d", ans - 1);
 
    return 0;
}
### Prompt

Develop a solution in Cpp 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<bits/stdc++.h>
const int MAXN = 2e5 + 10;
using namespace std;
int n;
int ans=0,sum=0;
int A[MAXN], B[MAXN], ta[MAXN], tb[MAXN], sa, sb, tot = 0, Do[MAXN], fa[MAXN];
int find(int x) {
    return fa[x] == x ? fa[x] : fa[x] = find(fa[x]);
}
int unionn(int x, int y) {
    fa[x] = y;
}
int main() {
    scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%d",&A[i]),sa^=A[i];A[n+1]=sa;
    for(int i=1;i<=n;i++)scanf("%d",&B[i]),sb^=B[i];B[n+1]=sb;
    memcpy(ta,A,sizeof(A));memcpy(tb,B,sizeof(B));
    n++;sort(ta+1,ta+n+1);sort(tb+1,tb+n+1);
    for(int i=1;i<n;i++)if(ta[i]!=tb[i])return puts("-1"), 0;
    for(int i=1;i<=n;i++)
        if(A[i]!=B[i]||(i==n)){
            Do[++sum]=A[i]; 
			Do[++sum]=B[i];
            if(i<n)ans++;
        }
    if(ans==0)return puts("0"), 0;
    sort(Do+1,Do+sum+1);
    sum=unique(Do+1,Do+sum+1)-Do-1;
    for(int i=1;i<=sum;i++)fa[i]=i;
    for(int i=1;i<=n;i++)
        if(A[i]!=B[i]) {
            A[i]=lower_bound(Do+1,Do+sum+1,A[i])-Do;
            B[i]=lower_bound(Do+1,Do+sum+1,B[i])-Do;
            unionn(find(A[i]),find(B[i]));
        }
    for(int i=1;i<=sum;i++)ans+=(fa[i]==i);
    printf("%d", ans - 1);
 
    return 0;
}
```