Problem:
Score : 600 points

Problem Statement

Given are an integer N and arrays S, T, U, and V, each of length N.
Construct an N×N matrix a that satisfy the following conditions:

a_{i,j} is an integer.

0 \leq a_{i,j} \lt 2^{64}.

If S_{i} = 0, the bitwise AND of the elements in the i-th row is U_{i}.

If S_{i} = 1, the bitwise OR of the elements in the i-th row is U_{i}.

If T_{i} = 0, the bitwise AND of the elements in the i-th column is V_{i}.

If T_{i} = 1, the bitwise OR of the elements in the i-th column is V_{i}.

However, there may be cases where no matrix satisfies the conditions.

Constraints

All values in input are integers.

1 \leq N \leq 500

0 \leq S_{i} \leq 1

0 \leq T_{i} \leq 1

0 \leq U_{i} \lt 2^{64}

0 \leq V_{i} \lt 2^{64}

Input

Input is given from Standard Input in the following format:

N
S_{1} S_{2} ... S_{N}
T_{1} T_{2} ... T_{N}
U_{1} U_{2} ... U_{N}
V_{1} V_{2} ... V_{N}

Output

If there exists a matrix that satisfies the conditions, print one such matrix in the following format:

a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}

Note that any matrix satisfying the conditions is accepted.

If no matrix satisfies the conditions, print -1.

Sample Input 1

2
0 1
1 0
1 1
1 0

Sample Output 1

1 1
1 0

In Sample Input 1, we need to find a matrix such that:

the bitwise AND of the elements in the 1-st row is 1;

the bitwise OR of the elements in the 2-nd row is 1;

the bitwise OR of the elements in the 1-st column is 1;

the bitwise AND of the elements in the 2-nd column is 0.

Sample Input 2

2
1 1
1 0
15 15
15 11

Sample Output 2

15 11
15 11

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;
using ll = unsigned long long;

const int MAX { 510 };

ll A[MAX][MAX];

int solve(int N, const vector<ll>& s, const vector<ll>& t, const vector<ll>& u, const vector<ll>& v) 
{
    vector<int> rs, cs;

    for (int i = 0; i < N; ++i)
    {
        if (s[i] == 0)
        {
            for (int j = 0; j < N; ++j)
                A[i][j] |= u[i];

            rs.push_back(i);
        }
    }
#ifdef DEBUG
    cout << "Apos S == 0\n";
    for (int i = 0; i < N; ++i)
        for (int j = 0; j < N; ++j)
            cout << A[i][j] << (j + 1 == N ? '\n' : ' ');
#endif
 
    for (int i = 0; i < N; ++i)
    {
        if (t[i] == 0)
        {
            for (int j = 0; j < N; ++j)
                A[j][i] |= v[i];

            cs.push_back(i);
        }
    }
#ifdef DEBUG
    cout << "Apos T == 0\n";
    for (int i = 0; i < N; ++i)
        for (int j = 0; j < N; ++j)
            cout << A[i][j] << (j + 1 == N ? '\n' : ' ');
#endif


    for (int i = 0; i < N; ++i)
    {
        if (s[i] == 0)
            continue;

        ll ans = A[i][0];

        for (int j = 1; j < N; ++j)
            ans |= A[i][j];
 
        if (ans == u[i])
            continue;

        bool ok = false;

        for (int j = 0; j < N; ++j)
            if (A[i][j] == 0)
            {
                A[i][j] = u[i];
                ok = true;
                break;
            }

        if (not ok)
            return -1;
    }

#ifdef DEBUG
    cout << "Apos S == 1\n";
    for (int i = 0; i < N; ++i)
        for (int j = 0; j < N; ++j)
            cout << A[i][j] << (j + 1 == N ? '\n' : ' ');
#endif

    for (int j = 0; j < N; ++j)
    {
        if (t[j] == 0)
            continue;

        ll ans = A[0][j];

        for (int i = 1; i < N; ++i)
            ans |= A[i][j];
 
        if (ans == v[j])
            continue;

        bool ok = false;

        for (int i = 0; i < N; ++i)
            if (A[i][j] == 0)
            {
                A[i][j] = v[j];
                ok = true;
                break;
            }

        if (not ok)
            return -1;
    }

#ifdef DEBUG
    cout << "Apos T == 1\n";
    for (int i = 0; i < N; ++i)
        for (int j = 0; j < N; ++j)
            cout << A[i][j] << (j + 1 == N ? '\n' : ' ');
#endif

    for (int i = 0; i < N; ++i)
    {
        ll ans = A[i][0];

        for (int j = 1; j < N; ++j)
        {
            if (s[i] == 0)
                ans &= A[i][j];
            else
                ans |= A[i][j];
        }

        if (ans != u[i])
            return -1;
    }

    for (int j = 0; j < N; ++j)
    {
        ll ans = A[0][j];

        for (int i = 1; i < N; ++i)
        {
            if (t[i] == 0)
                ans &= A[i][j];
            else
                ans |= A[i][j];
        }

        if (ans != v[j])
            return -1;
    }

    return 0;
}

int main()
{
    ios::sync_with_stdio(false);

    int N;
    cin >> N;

    vector<ll> s(N), t(N), u(N), v(N);

    for (int i = 0; i < N; ++i)
        cin >> s[i];

    for (int i = 0; i < N; ++i)
        cin >> t[i];

    for (int i = 0; i < N; ++i)
        cin >> u[i];

    for (int i = 0; i < N; ++i)
        cin >> v[i];

    auto ans = solve(N, s, t, u, v);

    if (ans == -1)
        cout << ans << '\n';
    else
    {
        for (int i = 0; i < N; ++i)
            for (int j = 0; j < N; ++j)
                cout << A[i][j] << (j + 1 == N ? '\n' : ' ');
    }

    return 0;
}