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 C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#include <regex>
using namespace std;
#define endl "\n";
typedef long long int ll;
ll P=1000000007;
ll MAX=P*(P-1);
ll MIN=-1*MAX;
const int sizes=200001;
int main()
{
    ll t,s,i,j,k,l,m,n,o,p,q,r,a[sizes],b[sizes],c[sizes],d[sizes],mat[1000][1000];
    cin>>n;
    for(i=0;i<n;i++)
    {
        cin>>a[i];
    }
    for(i=0;i<n;i++)
    cin>>b[i];
    for(i=0;i<n;i++)
    cin>>c[i];
    for(i=0;i<n;i++)
    cin>>d[i];
    ll flag=0;
    for(i=0;i<n;i++)
    {
        if(a[i]==0)
        {
            
            for(j=0;j<n;j++)
            {
                 l=c[i];
                 k=d[j];
                if(b[j]==0)
                {
                   
                    if((l&k)!=min(l,k))
                    flag=1;
                    mat[i][j]=max(l,k);
                }
                else 
                {
                    if((l&k)!=min(l,k))
                    flag=1;
                       mat[i][j]=k;
                }
            }
        }
        else 
        {
            ll x=0;
            for(j=0;j<n;j++)
            {
                l=c[i];
                k=d[j];
                if(b[j]==0)
                {
                    if(x!=l)
                    mat[i][j]=(l&k);
                    else 
                    mat[i][j]=k;
                    x|=mat[i][j];
                }
                else 
                {
                    if((l&k)!=min(l,k))
                    flag=1;
                    mat[i][j]=l;
                    if(x!=l)
                    mat[i][j]=(l&k);
                    else 
                    mat[i][j]=k;
                    x|=mat[i][j];
                }
            }
        }
    }
      if(flag)
      {
          cout<<"-1\n";
      }
      else 
      {
          for(i=0;i<n;i++)
          {
              for(j=0;j<n;j++)
              cout<<mat[i][j]<<" ";
              cout<<endl;
          }
      }
    return 0;
}