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 Rust (1.42.0) (judge verdict: Wrong Answer)

use proconio::*;

#[fastout]
fn run() {
    input! {
        n: usize,
        s: [usize; n],
        t: [usize; n],
        u: [usize; n],
        v: [usize; n],
    }
    let mut ans = vec![vec![0; n]; n];
    for k in 0..64 {
        let mut tmp = vec![vec![2; n]; n];
        let mut row = vec![];
        for (i, (&s, &u)) in s.iter().zip(u.iter()).enumerate() {
            let u = u >> k & 1;
            if s ^ u == 0 {
                row.push((i, u));
            } else {
                for t in tmp[i].iter_mut() {
                    *t = u;
                }
            }
        }
        let mut col = vec![];
        for (j, (&t, &v)) in t.iter().zip(v.iter()).enumerate() {
            let v = v >> k & 1;
            if t ^ v == 0 {
                col.push((j, v));
            } else {
                for i in 0..n {
                    if tmp[i][j] != 2 && tmp[i][j] ^ v != 0 {
                        println!("-1");
                        return;
                    }
                    tmp[i][j] = v;
                }
            }
        }
        if row.len() * col.len() == 0 {
        } else if row.len() == 1 {
            let (r, b) = row[0];
            for &(c, b) in col.iter() {
                tmp[r][c] = b;
            }
            if let Some(k) = col.iter().position(|&(c, b)| (0..n).any(|r| tmp[r][c] == b)) {
                tmp[r][col[k].0] = b;
            } else if tmp[r].iter().all(|x| *x != b) {
                println!("-1");
                return;
            }
        } else if col.len() == 1 {
            let (c, b) = col[0];
            for &(r, b) in row.iter() {
                tmp[r][c] = b;
            }
            if let Some(k) = row
                .iter()
                .position(|&(r, b)| (0..n).any(|c| tmp[r][c] == b))
            {
                tmp[row[k].0][c] = b;
            } else if (0..n).all(|x| tmp[x][c] != b) {
                println!("-1");
                return;
            }
        } else {
            for (i, &(r, _)) in row.iter().enumerate() {
                for (j, &(c, _)) in col.iter().enumerate() {
                    tmp[r][c] = (i ^ j) & 1;
                }
            }
        }
        for (a, t) in ans.iter_mut().zip(tmp.iter()) {
            for (a, t) in a.iter_mut().zip(t.iter()) {
                *a |= *t << k;
            }
        }
    }
    for (s, u) in [(s, u), (t, v)].iter() {
        for (ans, (&s, &u)) in ans.iter().zip(s.iter().zip(u.iter())) {
            if s == 0 {
                if ans.iter().fold(std::usize::MAX, |s, a| s & *a) != u {
                    println!("-1");
                    return;
                }
            } else {
                if ans.iter().fold(0, |s, a| s | *a) != u {
                    println!("-1");
                    return;
                }
            }
        }
        for i in 1..n {
            for j in 0..i {
                let x = ans[i][j];
                ans[i][j] = ans[j][i];
                ans[j][i] = x;
            }
        }
    }
    for a in ans {
        for a in a {
            print!("{} ", a);
        }
        println!();
    }
}

fn main() {
    run();
}