#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, a[N], b[N], ans;
map<int, int> rd, fa;
inline int Ask(int x) {
    return fa.count(x) ? fa[x] = Ask(fa[x]) : x;
}
inline void Merge(int x, int y) {
    x = Ask(x); y = Ask(y);
    if (x == y) return;
    fa[x] = y;
}
int main() {
    scanf("%d", &n);
    int a_xor = 0;
    for (int i = 1; i <= n; ++i) {
        scanf("%d", a + i);
        a_xor ^= a[i];
    }
    for (int i = 1; i <= n; ++i) {
        scanf("%d", b + i);
        if (a[i] != b[i]) {
            ++rd[a[i]];
            --rd[b[i]];
            Merge(a[i], b[i]);
            ++ans;
        }
    }
    if (!ans) return puts("0"), 0;
    int non_zero_cnt = 0, rt = Ask(a_xor), con_num = 0;
    for (map<int, int>::iterator it = rd.begin(); it != rd.end(); ++it) {
        if (Ask(it->first) == rt) non_zero_cnt += bool(it->second);
        else if (it->second) return puts("-1"), 0;
        con_num += !fa.count(it->first);
    }
    if (!rd.count(a_xor)) {
        printf("%d\n", ans + con_num);
    } else {
        if ((rd[a_xor] == 0 && non_zero_cnt == 0) || (rd[a_xor] == -1 && non_zero_cnt == 2)) {
            printf("%d\n", ans + con_num - 1);
        } else {
            puts("-1");
        }
    }
    return 0;
}
### Prompt

Your task is to create 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;
const int N = 1e5 + 10;
int n, a[N], b[N], ans;
map<int, int> rd, fa;
inline int Ask(int x) {
    return fa.count(x) ? fa[x] = Ask(fa[x]) : x;
}
inline void Merge(int x, int y) {
    x = Ask(x); y = Ask(y);
    if (x == y) return;
    fa[x] = y;
}
int main() {
    scanf("%d", &n);
    int a_xor = 0;
    for (int i = 1; i <= n; ++i) {
        scanf("%d", a + i);
        a_xor ^= a[i];
    }
    for (int i = 1; i <= n; ++i) {
        scanf("%d", b + i);
        if (a[i] != b[i]) {
            ++rd[a[i]];
            --rd[b[i]];
            Merge(a[i], b[i]);
            ++ans;
        }
    }
    if (!ans) return puts("0"), 0;
    int non_zero_cnt = 0, rt = Ask(a_xor), con_num = 0;
    for (map<int, int>::iterator it = rd.begin(); it != rd.end(); ++it) {
        if (Ask(it->first) == rt) non_zero_cnt += bool(it->second);
        else if (it->second) return puts("-1"), 0;
        con_num += !fa.count(it->first);
    }
    if (!rd.count(a_xor)) {
        printf("%d\n", ans + con_num);
    } else {
        if ((rd[a_xor] == 0 && non_zero_cnt == 0) || (rd[a_xor] == -1 && non_zero_cnt == 2)) {
            printf("%d\n", ans + con_num - 1);
        } else {
            puts("-1");
        }
    }
    return 0;
}
```