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<algorithm>
#include<string>
#include<vector>
#include<cmath>
//#include<bits/stdc++.h>
#define int long long
#define mod 1000000007
#define for0(i, n) for(int i = 0; i < (n); i++)
#define for1(i, n) for(int i = 1; i <= (n);i++)
#define mp make_pair
#define all(x) x.begin(),x.end()
#define puts(x) cout << x << endl
using namespace std;
int input() {
	int r;
	cin >> r;
	//scanf("%lld", &r);
	return r;
}
int a[510][510][70], n, s[510], t[510], u[510][70], v[510][70];
bool bu[510][70], bv[510][70];
int ansk[70];
signed main() {
	for1(k, 64)ansk[k] = (k == 1 ? 1 : ansk[k - 1] * 2);
	cin >> n;
	for1(i, n)cin >> s[i];
	for1(j, n)cin >> t[j];
	for1(i, n) {
		int ui; cin >> ui;
		for1(k, 64) {
			u[i][k] = ui % 2; ui /= 2;
		}
	}
	for1(j, n) {
		int vj; cin >> vj;
		for1(k, 64) {
			v[j][k] = vj % 2; vj /= 2;
		}
	}

	for1(i, n) for1(j, n) for1(k, 64) {
		a[i][j][k] = -1;
		if (s[i] == 0 && u[i][k] == 1) {
			if (t[j] == 1 && v[j][k] == 0) {
				cout << "-1\n"; return 0;
			}
			a[i][j][k] = 1;
		}
		if (s[i] == 1 && u[i][k] == 0) {
			if (t[j] == 0 && v[j][k] == 1) {
				cout << "-1\n"; return 0;
			}
			a[i][j][k] = 0;
		}
		if (t[j] == 1 && v[j][k] == 0)a[i][j][k] = 0;
		if (t[j] == 0 && v[j][k] == 1)a[i][j][k] = 1;
		if (a[i][j][k] == u[i][k])bu[i][k] = 1;
		if (a[i][j][k] == v[j][k])bv[j][k] = 1;
		if (a[i][j][k] >= 0)continue;
		if (u[i][k] == v[j][k]) {
			a[i][j][k] = u[i][k];
			bu[i][k] = 1;
			bv[j][k] = 1;
		}
	}
	for1(h, 50)for1(i, n) for1(j, n) for1(k, 64)if (a[i][j][k] == -1) {
		if (bu[i][k] && bv[j][k])a[i][j][k] = 0;
		else if (!bu[i][k] && bv[j][k]) {
			a[i][j][k] = u[i][k]; bu[i][k] = 1;
		}
		else if (bu[i][k] && !bv[j][k]) {
			a[i][j][k] = v[j][k]; bv[j][k] = 1;
		}
	}
	for1(i, n) for1(j, n) for1(k, 64)if (a[i][j][k] == -1) {
		cout << "-1\n"; return 0;
	}
	for1(i, n)for1(j, n) {
		int ans = 0;
		for1(k, 64)ans += ansk[k] * a[i][j][k];
		cout << ans << (j == n ? "\n" : " ");
	}
}