#include <bits/stdc++.h>
using namespace std;

const int maxn = 100010;
int n, x, ans, tot, a[maxn], b[maxn], c[maxn], d[maxn], fa[maxn];
map<int, int> mp;
bool chk[maxn];

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

int main() {
    scanf("%d", &n), n++;
    for (int i = 1; i < n; i++) {
        scanf("%d", &a[i]), c[i] = a[i], x ^= a[i];
    }
    a[n] = c[n] = x, sort(c + 1, c + n + 1), x = 0;
    for (int i = 1; i < n; i++) {
        scanf("%d", &b[i]), d[i] = b[i];
        fa[i] = i, x ^= b[i];
    }
    b[n] = d[n] = x, sort(d + 1, d + n + 1), fa[n] = n;
    int tot = 0;
    for (int i = 1; i <= n; i++) {
        if (c[i] ^ d[i]) printf("-1\n"), exit(0);
        if (!mp.count(c[i])) mp[c[i]] = ++tot;
    }
    for (int i = 1; i <= n; i++) {
        a[i] = mp[a[i]], b[i] = mp[b[i]];
        if (a[i] == b[i]) continue;
        if (i < n) ans++;
        int u = find(a[i]), v = find(b[i]);
        fa[u] = v, chk[u] = chk[v] = 1;
    }
    if (!ans) printf("0\n"), exit(0);
    chk[a[n]] = chk[b[n]] = 1;
    for (int i = 1; i <= tot; i++) {
        ans += (chk[i] && find(i) == i);
    }
    printf("%d\n", ans - 1);
    return 0;
}
### Prompt

In CPP, your task is to solve 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 maxn = 100010;
int n, x, ans, tot, a[maxn], b[maxn], c[maxn], d[maxn], fa[maxn];
map<int, int> mp;
bool chk[maxn];

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

int main() {
    scanf("%d", &n), n++;
    for (int i = 1; i < n; i++) {
        scanf("%d", &a[i]), c[i] = a[i], x ^= a[i];
    }
    a[n] = c[n] = x, sort(c + 1, c + n + 1), x = 0;
    for (int i = 1; i < n; i++) {
        scanf("%d", &b[i]), d[i] = b[i];
        fa[i] = i, x ^= b[i];
    }
    b[n] = d[n] = x, sort(d + 1, d + n + 1), fa[n] = n;
    int tot = 0;
    for (int i = 1; i <= n; i++) {
        if (c[i] ^ d[i]) printf("-1\n"), exit(0);
        if (!mp.count(c[i])) mp[c[i]] = ++tot;
    }
    for (int i = 1; i <= n; i++) {
        a[i] = mp[a[i]], b[i] = mp[b[i]];
        if (a[i] == b[i]) continue;
        if (i < n) ans++;
        int u = find(a[i]), v = find(b[i]);
        fa[u] = v, chk[u] = chk[v] = 1;
    }
    if (!ans) printf("0\n"), exit(0);
    chk[a[n]] = chk[b[n]] = 1;
    for (int i = 1; i <= tot; i++) {
        ans += (chk[i] && find(i) == i);
    }
    printf("%d\n", ans - 1);
    return 0;
}
```