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)

#[allow(unused_imports)]
use proconio::{
    fastout, input,
    marker::{Bytes, Chars, Isize1, Usize1},
};

#[allow(unused_macros)]
macro_rules! echo {
    ($($e:expr),+) => ( { $(println!("{}", $e))+ } );
}

fn main() {
    input! {
        n: usize,
        ss: [usize; n],
        ts: [usize; n],
        us: [usize; n],
        vs: [usize; n],
    }

    let mut ans = vec![vec![None; n]; n];

    for (i, &s) in ss.iter().enumerate() {
        let u = us[i];

        if s == 0 {
            for e in ans[i].iter_mut() {
                *e = Some(u);
            }
        }
    }

    for (j, &t) in ts.iter().enumerate() {
        let v = vs[j];

        if t == 0 {
            for es in ans.iter_mut() {
                match es[j] {
                    Some(u) => {
                        es[j] = Some(u | v);
                    }
                    None => {
                        es[j] = Some(v);
                    }
                }
            }
        }
    }

    for (i, &s) in ss.iter().enumerate() {
        let u = us[i];

        if s == 1 {
            for (j, e) in ans[i].iter_mut().enumerate() {
                match *e {
                    Some(v) => {
                        if (u | v) != u {
                            echo!(-1);
                            return;
                        }
                    }
                    None => {
                        let v = vs[j];
                        *e = Some(u & v);
                    }
                }
            }
        }
    }

    for (j, &t) in ts.iter().enumerate() {
        let v = vs[j];

        if t == 1 {
            for (i, es) in ans.iter_mut().enumerate() {
                match es[j] {
                    Some(u) => {
                        if (u | v) != v {
                            echo!(-1);
                            return;
                        }
                    }
                    None => {
                        let u = us[i];
                        es[j] = Some(u & v);
                    }
                }
            }
        }
    }

    for es in ans.iter() {
        for (i, e) in es.iter().enumerate() {
            print!("{}", e.unwrap());
            if i != es.len() - 1 {
                print!(" ");
            }
        }
        println!("");
    }
}