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<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<stack>
#include<vector>
#include<map>
#define int long long
using namespace std;
const int maxn = 5e2+10;

int read(){
	int x=0,f=1;
	char c=getchar();
	while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}
	while(isdigit(c)){x=(x<<3)+(x<<1)+c-48;c=getchar();}
	return x*f;
}

int k,m,n;
int a[maxn],b[maxn],c[maxn],d[maxn],ans[maxn][maxn],anss[maxn][maxn];
int check1[maxn],check2[maxn];
vector<int> a1,a2;
int P[100];
inline void solve(int x){
	memset(check1,0,sizeof(check1));
	memset(check2,0,sizeof(check2));
	memset(anss,0x3f,sizeof(anss));a1.clear();a2.clear();
	for(register int i=1;i<=n;i++){
		if(a[i]==1 && c[i]%2==0){
			check1[i]=1;
			for(register int j=1;j<=n;j++)anss[i][j]=0;
		}
		if(a[i]==0 && c[i]%2==1){
			check1[i]=1;
			for(register int j=1;j<=n;j++)anss[i][j]=1;
		}
	}
	for(register int i=1;i<=n;i++){
		if(b[i]==1 && d[i]%2==0){
			check2[i]=1;
			for(register int j=1;j<=n;j++){if(anss[j][i]==1){puts("-1");exit(0);}anss[j][i]=0;}
		}
		if(b[i]==0 && d[i]%2==1){
			check2[i]=1;
			for(register int j=1;j<=n;j++){if(anss[j][i]==0){puts("-1");exit(0);}anss[j][i]=1;}
		}
	}
	for(register int i=1;i<=n;i++){
		for(register int j=1;j<=n;j++){
			if(a[i]-c[i]%2==0 && b[j]-d[j]%2==0 && a[i]==1 && b[j]==1){
				check1[i]=1;check2[j]=1;
				anss[i][j]=1;
			}
			if(a[i]-c[i]%2==0 && b[j]-d[j]%2==0 && a[i]==0 && b[j]==0){
				check1[i]=1;check2[j]=1;
				anss[i][j]=0;
			}
		}
	}
	int num1=0,num2=0;
	for(register int i=1;i<=n;i++)if(check1[i]==-1){num1++;a1.push_back(i);}
	for(register int i=1;i<=n;i++)if(check2[i]==-1){num2++;a2.push_back(i);}
	if(num1!=0 && num2!=0){
		if(num1==1 || num2==1){puts("-1");exit(0);}
		int flag=1;
		if(num1<=num2 && a1[0]==0)flag=0;
		else flag=1;
		for(register int i=0;i<=num1;i++){
			for(register int j=0;j<=num2;j++){
				if(i==j)anss[i][j]=flag;
				else anss[i][j]=flag^1;
			}
		}
	}
	for(register int i=1;i<=n;i++)
		for(register int j=1;j<=n;j++)
			if(anss[i][j]!=0 && anss[i][j]!=1)anss[i][j]=0;
	for(register int i=1;i<=n;i++)
		for(register int j=1;j<=n;j++){
			ans[i][j]+=anss[i][j]*P[x];
		}
//	for(register int i=1;i<=n;i++){
//		for(register int j=1;j<=n;j++){
//			cout<<anss[i][j]<<" ";
//		}
//		puts("");
//	}
	for(register int i=1;i<=n;i++){c[i]/=2;d[i]/=2;}
//	for(register int i=1;i<=n;i++)cout<<c[i]<<" "<<d[i]<<endl;puts("");
}

signed main(){
	P[0]=1;
	for(register int i=1;i<=65;i++){
		P[i]=P[i-1]*2ll;
	}
	n=read();
	for(register int i=1;i<=n;i++)a[i]=read();
	for(register int i=1;i<=n;i++)b[i]=read();
	for(register int i=1;i<=n;i++)c[i]=read();
	for(register int i=1;i<=n;i++)d[i]=read();
	int now=1;
	for(register int i=0;i<=63;i++)solve(i);
	for(register int i=1;i<=n;i++){
		for(register int j=1;j<=n;j++)
			printf("%lld ",ans[i][j]);
		puts("");
	}
	return 0;
}