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: Accepted)

#include <bits/stdc++.h>
using namespace std;
using ull = unsigned long long;

int n;
bool s[500], t[500], u[500], v[500];
int a[500][500];
ull ans[500][500];
void end(){
    cout << -1 << endl;
    exit(0);
}
void solve(){
    memset(a, 0xff, sizeof(a));
    vector<int> x(n), y(n);
    auto px = x.begin(), py = y.begin();
    for(int i = 0; i < n; i++) x[i] = y[i] = i;
    for(int i = 0; i < n; i++){
        if(s[i] ^ u[i]){
            for(int j = 0; j < n; j++) a[i][j] = u[i];
            px = x.erase(px);
        }
        else px++;
    }
    for(int j = 0; j < n; j++){
        if(t[j] ^ v[j]){
            for(int i = 0; i < n; i++){
                if(a[i][j] == (v[j] ^ 1)) end();
                a[i][j] = v[j];
            }
            py = y.erase(py);
        }
        else py++;
    }
    if(x.size() > 1 && y.size() > 1){
        bool b = 0;
        for(int i : x){
            bool c = b;
            for(int j : y){
                a[i][j] = c;
                c = !c;
            }
            b = !b;
        }
        return;
    }
    if(x.size() == 1){
        const int i = x[0];
        for(int j : y){
            a[i][j] = u[i];
            bool flag = 1;
            for(int i = 0; i < n; i++) if(a[i][j] == v[j]){
                flag = 0;
                break;
            }
            if(flag) a[i][j] = v[j];
        }
    }
    if(y.size() == 1){
        const int j = y[0];
        for(int i : x){
            a[i][j] = v[j];
            bool flag = 1;
            for(int j = 0; j < n; j++) if(a[i][j] == u[i]){
                flag = 0;
                break;
            }
            if(flag) a[i][j] = u[i];
        }
    }
    for(int i : x){
        bool flag = 1;
        for(int j = 0; j < n; j++) if(a[i][j] == u[i]){
            flag = 0;
            break;
        }
        if(flag) end();
    }
    for(int j : y){
        bool flag = 1;
        for(int i = 0; i < n; i++) if(a[i][j] == v[j]){
            flag = 0;
            break;
        }
        if(flag) end();
    }
}
int main(){
    cin >> n;
    vector<ull> u_(n), v_(n);
    for(int i = 0; i < n; i++) cin >> s[i];
    for(int i = 0; i < n; i++) cin >> t[i];
    for(auto& i : u_) cin >> i;
    for(auto& i : v_) cin >> i;
    for(int bit = 64; bit--; ){
        for(int i = 0; i < n; i++) u[i] = u_[i] >> bit & 1;
        for(int i = 0; i < n; i++) v[i] = v_[i] >> bit & 1;
        solve();
        for(int i = 0; i < n; i++) for(int j = 0; j < n; j++){
            ans[i][j] <<= 1;
            ans[i][j] |= a[i][j];
        }
    }
    for(int i = 0; i < n; i++){
        for(int j = 0; j < n; j++) cout << ans[i][j] << ' ';
        cout << '\n';
    }
}