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 itertools::join;
fn main() {
    proconio::input! {
        n: usize,
        s: [u8;n],
        t: [u8;n],
        u: [u64;n],
        v: [u64;n],
    }
    let mut m = vec![vec![0;n];n];
    for i in 0..n { for j in 0..n {
        let mut a = u[i]&v[j];
        if s[i]==0 {a |= u[i];}
        if t[j]==0 {a |= v[j];}
        if s[i]!=0 {a &= u[i];}
        if t[j]!=0 {a &= v[j];}
        m[i][j] = a;
    }}
    for a in &m {
        eprintln!("{}",join(a.iter(), " "));
    }
    if (0..n).any(|k|
        u[k] != if s[k]!=0 {
            (0..n).fold(0, |a,j| a|m[k][j])
        }
        else {
            (0..n).fold(std::u64::MAX, |a,j| a&m[k][j])
        }
        ||
        v[k] != if t[k]!=0 {
            (0..n).fold(0, |a,i| a|m[i][k])
        }
        else {
            (0..n).fold(std::u64::MAX, |a,i| a&m[i][k])
        }
    ) {
        println!("{}",-1);
        return;
    }
    for a in m {
        println!("{}",join(a.iter(), " "));
    }
}