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: Accepted)

use itertools::Itertools as _;
use proconio::*;
macro_rules! p {
    ($x:expr) => {
        println!("{}", $x);
    };
}
fn main() {
    input! {
        n:usize,
        s:[usize;n],
        t:[usize;n],
        u:[u64;n],
        v:[u64;n],
    }
    let mask = std::u64::MAX;
    let (or, and) = (1, 0);
    // 1確定
    let mut one = vec![vec![0u64; n]; n];
    // 0確定
    let mut zero = vec![vec![0u64; n]; n];

    // 自明
    for i in 0..n {
        for j in 0..n {
            if s[i] == and {
                one[i][j] |= u[i];
            } else {
                zero[i][j] |= !u[i];
            }
            if t[j] == and {
                one[i][j] |= v[j];
            } else {
                zero[i][j] |= !v[j];
            }
            if one[i][j] & zero[i][j] != 0 {
                p!(-1);
                return;
            }
        }
    }
    let mut ans = one.clone();

    // 条件を満たさない行/列のあるbitを調べる
    let check = |ans: &[Vec<u64>]| {
        let mut u_check = (0..n)
            .map(|i| if s[i] == or { 0 } else { mask })
            .collect_vec();
        let mut v_check = (0..n)
            .map(|j| if t[j] == or { 0 } else { mask })
            .collect_vec();

        for i in 0..n {
            for j in 0..n {
                if s[i] == and {
                    u_check[i] &= ans[i][j];
                } else {
                    u_check[i] |= ans[i][j];
                }
                if t[j] == and {
                    v_check[j] &= ans[i][j];
                } else {
                    v_check[j] |= ans[i][j];
                }
            }
        }
        let check = u
            .iter()
            .zip(&u_check)
            .chain(v.iter().zip(&v_check))
            .map(|(x, y)| x ^ y)
            .fold(0, |x, y| x | y);
        check
    };

    // 市松
    // 未確定が2*2以上なら必ず通る
    // 市松が間違っていたので修正
    let mut iflip;
    let mut jflip = 0u64;
    for i in 0..n {
        iflip = 0;
        jflip ^= if s[i] == and { !u[i] } else { u[i] };
        for j in 0..n {
            ans[i][j] |= !zero[i][j] & (iflip ^ jflip);
            iflip ^= !(one[i][j] | zero[i][j]);
        }
    }

    let fail = check(&ans);
    // 全部or
    for i in 0..n {
        for j in 0..n {
            ans[i][j] &= !fail;
            ans[i][j] |= fail & (u[i] | v[j]) & !zero[i][j];
        }
    }

    let fail = check(&ans);
    // 全部and
    for i in 0..n {
        for j in 0..n {
            ans[i][j] &= !fail;
            ans[i][j] |= fail & ((u[i] & v[j]) | one[i][j]);
        }
    }

    let fail = check(&ans);
    if fail != 0 {
        p!(-1);
        return;
    }

    ans.iter().for_each(|x| p!(x.iter().join(" ")));
}