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>

using namespace std;

int n;
short s[510], t[510];
unsigned long long u[510], v[510], a[64][510][510], ans[510][510], p[510];
int cntHor[510][64], cntVer[510][64];
bitset<64> hor[510], ver[510];

void gen(){
	p[0] = 1;

	for(int i = 1; i < 64; i++) p[i] = p[i - 1] << 1;
}

void Fail(){
	cout << "-1\n"; exit(0);
}

int main(){
	gen();
	ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
	
	cin >> n;

	for(int i = 1; i <= n; i++) cin >> s[i];
	for(int i = 1; i <= n; i++) cin >> t[i];
	for(int i = 1; i <= n; i++) cin >> u[i];
	for(int i = 1; i <= n; i++) cin >> v[i];

	for(int k = 0; k < 64; k++){
		for(int i = 1; i <= n; i++){
			for(int j = 1; j <= n; j++){
				if(s[i] == 0){
					if(u[i] & p[k]) a[k][i][j] = 1;
				} else {
					if(u[i] & p[k]);
					else a[k][i][j] = -1;
				}
			}
		}
	}

	for(int k = 0; k < 64; k++){
		for(int j = 1; j <= n; j++){
			for(int i = 1; i <= n; i++){
				if(t[j] == 0){
					if(v[j] & p[k]){
						if(a[k][i][j] == -1) Fail();

						a[k][i][j] = 1;
					}
				} else {
					if(v[j] & p[k]);
					else {
						if(a[k][i][j] == 1) Fail();
						a[k][i][j] = -1;
					}
				}
			}
		}
	}

	for(int k = 0; k < 64; k++){
		for(int i = 1; i <= n; i++){
			for(int j = 1; j <= n; j++){
				if(a[k][i][j] == 0){
					if(s[i] == 0) cntHor[i][k]++;
					if(t[j] == 0) cntVer[j][k]++;
				}

				if(a[k][i][j] == 1){
					if(s[i] == 1) hor[i][k] = true;
					if(t[j] == 1) ver[j][k] = true;
				}
			}
		}
	}

	for(int k = 0; k < 64; k++){
		for(int i = 1; i <= n; i++){
			for(int j = 1; j <= n; j++){
				if(a[k][i][j] == 0){
					if(s[i] == 1 && (t[j] == 1 || cntVer[j][k] > 1) && !hor[i][k]){
						cntVer[j][k]--;
						a[k][i][j] = 1;
						hor[i][k] = true;
					} else if (t[j] == 1 && (s[i] == 1 || cntHor[i][k] > 1) && !ver[j][k]){
						cntHor[i][k]--;
						a[k][i][j] = 1;
						ver[j][k] = true;
					}
				}

				if(a[k][i][j] == 1) ans[i][j] += p[k];
			}
		}
	}

	for(int i = 1; i <= n; i++){
		unsigned long long x, y;
		x = ans[i][1], y = ans[i][1];
 
		for(int j = 1; j <= n; j++){
			x &= ans[i][j];
			y |= ans[i][j];
		}
 
		if(s[i] == 0 && u[i] != x) Fail();
		else if (s[i] == 1 && u[i] != y) Fail();
	}
 
	for(int j = 1; j <= n; j++){
		unsigned long long x, y;
		x = ans[1][j], y = ans[1][j];
 
		for(int i = 1; i <= n; i++){
			x &= ans[i][j];
			y |= ans[i][j];
		}
 
		if(t[j] == 0 && v[j] != x) Fail();
		else if (t[j] == 1 && v[j] != y) Fail();
	}

	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= n; j++){
			cout << ans[i][j] << " \n"[j == n];
		}
	}

	return 0;
}