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 Python (3.8.2) (judge verdict: Accepted)

import numpy as np

N = int(input())

S = list(map(int,input().split()))
T = list(map(int,input().split()))
U = np.array(list(map(int,input().split())),dtype = np.uint64)
V = np.array(list(map(int,input().split())),dtype = np.uint64)

ans = np.zeros((N,N),dtype = np.uint64)

for n in range(64):
  u = (U>>n)&1
  v = (V>>n)&1
  su = [int(S[i]*2+u[i]) for i in range(N)]
  tv = [int(T[i]*2+v[i]) for i in range(N)]
  suzerozero = []
  suoneone = []
  tvzerozero = []
  tvoneone = []
  for i in range(N):
    if su[i] == 0:
      suzerozero.append(i)
    if su[i] == 3:
      suoneone.append(i)
  for i in range(N):
    if tv[i] == 0:
      tvzerozero.append(i)
    if tv[i] == 3:
      tvoneone.append(i)
  
  b = np.zeros((N,N),dtype = np.uint64)
  ###行のAND=1とOR=0を埋める###
  for i in range(N):
    if su[i] == 1:
      b[i] = np.ones(N,int)
  ###列のAND=1とOR=0を埋める###
  for i in range(N):
    if tv[i] == 1:
      b[:,i] = np.ones(N,int)
  ###行と列の値が一致するところを埋める###  
  b |= np.dot(np.array([u]).T,np.ones((1,N),dtype = np.uint64)) & np.dot(np.array([v]).T,np.ones((1,N),dtype = np.uint64)).T
  ###行のOR=1を満たしていく### 
  if 1 not in tv and 3 not in tv:
    m = len(tvzerozero)
    now = 0
    for i in suoneone:
      b[i][tvzerozero[now%m]] += 1
      now += 1
  ###列のOR=1を満たしていく### 
  if 1 not in su and 3 not in su:
    m = len(suzerozero)
    now = 0
    for i in tvoneone:
      b[suzerozero[now%m]][i] += 1
      now += 1
  ans += b<<n

def end():
  print(-1)
  exit()
for i in range(N):
  if S[i] == 0:
    now = ans[i][0]
    for j in range(1,N):
      now &= ans[i][j]
  if S[i] == 1:
    now = ans[i][0]
    for j in range(1,N):
      now |= ans[i][j]
  if now != U[i]:
    end()
for i in range(N):
  if T[i] == 0:
    now = ans[0][i]
    for j in range(1,N):
      now &= ans[j][i]
  if T[i] == 1:
    now = ans[0][i]
    for j in range(1,N):
      now |= ans[j][i]
  if now != V[i]:
    end()

for item in ans:
  print(*list(item))