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 Java (OpenJDK 11.0.6) (judge verdict: Runtime Error)


import java.util.*;

public class Main {
    private static final Scanner sc = new Scanner(System.in);
    private static final int M = 62;

    private long[] S;
    private long[] T;
    private long[] U;
    private long[] V;
    private int[][][] mat;
    private long[][] ans;
    private int N;

    private void read(long[] v, int N) {
        for (int i = 0; i < N; i++) {
            v[i] = sc.nextInt();
        }
    }

    private int bit(long v, int k) {
        return (v & (1L << k)) > 0 ? 1 : 0;
    }

    private boolean rowExist(int k, int r, int exclude, int value) {
        for (int c = 0; c < N; c++) {
            if (c != exclude && mat[k][r][c] == value) {
                return true;
            }
        }
        return false;
    }

    private boolean colExist(int k, int c, int exclude, int value) {
        for (int r = 0; r < N; r++) {
            if (r != exclude && mat[k][r][c] == value) {
                return true;
            }
        }
        return false;
    }

    private void fill(int k) {
        // AND1 row
        for (int r = 0; r < N; r++) {
            if (S[r] == 0 && bit(U[r], k) == 1) {
                for (int c = 0; c < N; c++) {
                    mat[k][r][c] = 1;
                }
            }
        }

        // AND1 col
        for (int c = 0; c < N; c++) {
            if (T[c] == 0 && bit(V[c], k) == 1) {
                for (int r = 0; r < N; r++) {
                    mat[k][r][c] = 1;
                }
            }
        }

        for (int r = 0; r < N; r++) {
            for (int c = 0; c < N; c++) {
                // OR1 OR1
                if (S[r] == 1 && bit(U[r], k) == 1 && T[c] == 1 && bit(V[c], k) == 1) {
                    mat[k][r][c] = 1;
                }

                // AND0 OR1
                if (S[r] == 0 && bit(U[r], k) == 0 && T[c] == 1 && bit(V[c], k) == 1) {
                    if (mat[k][r][c] == 0 && !colExist(k, c, r, 1) && rowExist(k, r, c, 0)) {
                        mat[k][r][c] = 1;
                    }
                }

                // OR1 AND0
                if (S[r] == 1 && bit(U[r], k) == 1 && T[c] == 0 && bit(V[c], k) == 0) {
                    if (mat[k][r][c] == 0 && !rowExist(k, r, c, 1) && colExist(k, c, r, 0)) {
                        mat[k][r][c] = 1;
                    }
                }
            }
        }
    }

    private boolean check() {
        for (int r = 0; r < N; r++) {
            long value = ans[r][0];
            for (int c = 1; c < N; c++) {
                value = S[r] == 0 ? (value & ans[r][c]) : (value | ans[r][c]);
            }
            if (value != U[r]) {
                return false;
            }
        }

        for (int c = 0; c < N; c++) {
            long value = ans[0][c];
            for (int r = 0; r < N; r++) {
                value = T[c] == 0 ? (value & ans[r][c]) : (value | ans[r][c]);
            }
            if (value != V[c]) {
                return false;
            }
        }
        return true;
    }

    private void solve() {
        N = sc.nextInt();
        S = new long[N]; T = new long[N]; U = new long[N]; V = new long[N];
        read(S, N); read(T, N); read(U, N); read(V, N);
        mat = new int[M][N][N];

        for (int k = 0; k < M; k++) {
            fill(k);
        }

        ans = new long[N][N];
        for (int r = 0; r < N; r++) {
            for (int c = 0; c < N; c++) {
                long value = 0;
                for (int k = M - 1; k >= 0; k--) {
                    value = (value << 1) + mat[k][r][c];
                }
                ans[r][c] = value;
            }
        }

        if (check()) {
            for (int r = 0; r < N; r++) {
                for (int c = 0; c < N; c++) {
                    System.out.print(ans[r][c]);
                    System.out.print(c == N - 1 ? "\n" : " ");
                }
            }
        } else {
            System.out.println("-1");
        }
    }

    public static void main(String[] args) {
        new Main().solve();
    }
}