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

map<int, int> counter(const vector<int>& xs)
{
    map<int, int> counts;
    for (const int x : xs)
        counts[x]++;
    return counts;
}

int main()
{
    int N;
    cin >> N;

    vector<int> a(N+1);
    for (int i = 0; i < N; i++)
    {
        cin >> a[i];
        a[N] ^= a[i];
    }

    vector<int> b(N+1);
    for (int i = 0; i < N; i++)
    {
        cin >> b[i];
        b[N] ^= b[i];
    }

    if (counter(a) != counter(b))
    {
        cout << -1 << endl;
        return 0;
    }

    int edges = 0;
    map<int, vector<int>> out_neighbors;
    for (int i = 0; i <= N; i++)
    {
        if (i < N && a[i] == b[i]) continue;
        edges++;
        out_neighbors[a[i]].push_back(b[i]);
    }

    set<int> seen;
    const function<void(int)> dfs = [&](const int v)
    {
        seen.insert(v);
        for (const int w : out_neighbors[v])
            if (!seen.count(w))
                dfs(w);
    };

    int components = 0;
    for (const auto kv : out_neighbors)
    {
        if (seen.count(kv.first)) continue;
        else components++;
        dfs(kv.first);
    }

    cout << edges + components - 2 << endl;
}

### 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;

map<int, int> counter(const vector<int>& xs)
{
    map<int, int> counts;
    for (const int x : xs)
        counts[x]++;
    return counts;
}

int main()
{
    int N;
    cin >> N;

    vector<int> a(N+1);
    for (int i = 0; i < N; i++)
    {
        cin >> a[i];
        a[N] ^= a[i];
    }

    vector<int> b(N+1);
    for (int i = 0; i < N; i++)
    {
        cin >> b[i];
        b[N] ^= b[i];
    }

    if (counter(a) != counter(b))
    {
        cout << -1 << endl;
        return 0;
    }

    int edges = 0;
    map<int, vector<int>> out_neighbors;
    for (int i = 0; i <= N; i++)
    {
        if (i < N && a[i] == b[i]) continue;
        edges++;
        out_neighbors[a[i]].push_back(b[i]);
    }

    set<int> seen;
    const function<void(int)> dfs = [&](const int v)
    {
        seen.insert(v);
        for (const int w : out_neighbors[v])
            if (!seen.count(w))
                dfs(w);
    };

    int components = 0;
    for (const auto kv : out_neighbors)
    {
        if (seen.count(kv.first)) continue;
        else components++;
        dfs(kv.first);
    }

    cout << edges + components - 2 << endl;
}

```