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 <iostream>
#include <vector>

using ulint = unsigned long long;

void fail() {
    std::cout << -1 << std::endl;
    std::exit(0);
}

void solve() {
    int n;
    std::cin >> n;

    std::vector<int> ss(n), ts(n);
    for (auto& s : ss) std::cin >> s;
    for (auto& t : ts) std::cin >> t;

    std::vector<ulint> us(n), vs(n);
    for (auto& u : us) std::cin >> u;
    for (auto& v : vs) std::cin >> v;

    std::vector<std::vector<ulint>> ans(n, std::vector<ulint>(n, 0));
    for (int k = 0; k < 64; ++k) {
        std::vector<int> xs(n), ys(n);
        for (int i = 0; i < n; ++i) {
            xs[i] = (us[i] >> k) & 1;
            ys[i] = (vs[i] >> k) & 1;
        }

        std::vector<bool> xok(n, false), yok(n, false);
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                ulint b;
                if ((ss[i] == 0 && xs[i] == 1) ||
                    (ts[j] == 0 && ys[j] == 1)) {
                    b = 1;
                } else if (
                    (ss[i] == 1 && xs[i] == 0) ||
                    (ts[j] == 1 && ys[j] == 0)) {
                    b = 0;
                } else if (!xok[i]) {
                    b = xs[i];
                    xok[i] = true;
                } else if (!yok[j]) {
                    b = ys[j];
                    yok[j] = true;
                } else {
                    b = 0;
                }

                ans[i][j] |= (b << k);
            }
        }
    }

    for (int i = 0; i < n; ++i) {
        if (ss[i] == 0) {
            ulint b = ulint(-1);
            for (int j = 0; j < n; ++j) b &= ans[i][j];
            if (b != us[i]) fail();
        } else {
            ulint b = 0;
            for (int j = 0; j < n; ++j) b |= ans[i][j];
            if (b != us[i]) fail();
        }
    }

    for (int j = 0; j < n; ++j) {
        if (ts[j] == 0) {
            ulint b = ulint(-1);
            for (int i = 0; i < n; ++i) b &= ans[i][j];
            if (b != vs[j]) fail();
        } else {
            ulint b = 0;
            for (int i = 0; i < n; ++i) b |= ans[i][j];
            if (b != vs[j]) fail();
        }
    }

    for (const auto& v : ans) {
        for (const auto& x : v) {
            std::cout << x << " ";
        }
        std::cout << std::endl;
    }
}

int main() {
    std::cin.tie(nullptr);
    std::ios::sync_with_stdio(false);

    solve();

    return 0;
}