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;
#define F first
#define S second
#define R cin>>
#define Z class
#define ll long long
#define ln cout<<'\n'
#define in(a) insert(a)
#define pb(a) push_back(a)
#define pd(a) printf("%.10f\n",a)
#define mem(a) memset(a,0,sizeof(a))
#define all(c) (c).begin(),(c).end()
#define iter(c) __typeof((c).begin())
#define rrep(i,n) for(ll i=(ll)(n)-1;i>=0;i--)
#define REP(i,m,n) for(ll i=(ll)(m);i<(ll)(n);i++)
#define rep(i,n) REP(i,0,n)
#define tr(it,c) for(iter(c) it=(c).begin();it!=(c).end();it++)
template<Z A>void pr(A a){cout<<a;ln;}
template<Z A,Z B>void pr(A a,B b){cout<<a<<' ';pr(b);}
template<Z A,Z B,Z C>void pr(A a,B b,C c){cout<<a<<' ';pr(b,c);}
template<Z A,Z B,Z C,Z D>void pr(A a,B b,C c,D d){cout<<a<<' ';pr(b,c,d);}
template<Z A>void PR(A a,ll n){rep(i,n){if(i)cout<<' ';cout<<a[i];}ln;}
ll check(ll n,ll m,ll x,ll y){return x>=0&&x<n&&y>=0&&y<m;}
const ll MAX=1e9+7,MAXL=1LL<<61,dx[8]={-1,0,1,0,-1,-1,1,1},dy[8]={0,1,0,-1,-1,1,1,-1};
typedef pair<ll,ll> P;

void Main() {
  ll n;
  R n;
  ll s[n],t[n],u[n],v[n];
  rep(i,n) R s[i];
  rep(i,n) R t[i];
  rep(i,n) R u[i];
  rep(i,n) R v[i];
  ll a[n][n];
  memset(a,-1,sizeof(a));
  rep(i,n) {
    if(!s[i]) {
      rep(j,n) {
        if(a[i][j]==-1) a[i][j]=0;
        a[i][j]|=u[i];
      }
    }
    if(!t[i]) {
      rep(j,n) {
        if(a[j][i]==-1) a[j][i]=0;
        a[j][i]|=v[i];
      }
    }
  }
  rep(i,n) {
    if(s[i]) {
      rep(j,n) {
        if(a[i][j]==-1) a[i][j]=u[i];
      }
    }
    if(t[i]) {
      rep(j,n) {
        if(a[j][i]==-1) a[j][i]=v[i];
      }
    }
  }
  rep(i,n) {
    ll x=0;
    if(!s[i]) {
      x=a[i][0];
      rep(j,n) x&=a[i][j];
    } else {
      rep(j,n) x|=a[i][j];
    }
    if(x!=u[i]) {
      pr(-1);
      return;
    }
  }
  rep(i,n) {
    ll x=0;
    if(!t[i]) {
      x=a[0][i];
      rep(j,n) x&=a[j][i];
    } else {
      rep(j,n) x|=a[j][i];
    }
    if(x!=v[i]) {
      pr(-1);
      return;
    }
  }
  rep(i,n) PR(a[i],n);
}

int main(){ios::sync_with_stdio(0);cin.tie(0);Main();return 0;}