Problem: 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.

Examples

Input

2
0 1
1 0
1 1
1 0


Output

1 1
1 0


Input

2
1 1
1 0
15 15
15 11


Output

15 11
15 11

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

#define AND 0
#define OR  1

bool check(vector<vector<int>> A,vector<int> s,vector<int> t,vector<int> u,vector<int> v){
	int n=A.size();
	rep(i,n){
		int z;
		if(s[i]==AND){
			z=1;
			rep(j,n) z&=A[i][j];
		}
		else{
			z=0;
			rep(j,n) z|=A[i][j];
		}
		if(z!=u[i]) return false;
	}
	rep(j,n){
		int z;
		if(t[j]==AND){
			z=1;
			rep(i,n) z&=A[i][j];
		}
		else{
			z=0;
			rep(i,n) z|=A[i][j];
		}
		if(z!=v[j]) return false;
	}
	return true;
}

pair<bool,vector<vector<int>>> solve(int n,vector<int> s,vector<int> t,vector<int> u,vector<int> v){
	vector<vector<int>> res(n,vector<int>(n,-1));

	vector<int> R,C;
	int s0=0,s1=0,t0=0,t1=0;
	rep(i,n){
		if(s[i]==AND && u[i]==1){
			rep(j,n) res[i][j]=1;
			s1++;
		}
		else if(s[i]==OR && u[i]==0){
			rep(j,n) res[i][j]=0;
			s0++;
		}
		else R.emplace_back(i);
	}
	rep(j,n){
		if(t[j]==AND && v[j]==1){
			rep(i,n) res[i][j]=1;
			t1++;
		}
		else if(t[j]==OR && v[j]==0){
			rep(i,n) res[i][j]=0;
			t0++;
		}
		else C.emplace_back(j);
	}
	if((s0>0 && t1>0) || (s1>0 && t0>0)) return {false,{}};

	if(R.size()==1){
		int x=u[R[0]];
		rep(j,C.size()){
			if((v[C[j]]==0 && s0>0) || (v[C[j]]==1 && s1>0)) res[R[0]][C[j]]=x;
			else                                             res[R[0]][C[j]]=v[C[j]];
		}
	}
	else if(C.size()==1){
		int x=v[C[0]];
		rep(i,R.size()){
			if((u[R[i]]==0 && t0>0) || (u[R[i]]==1 && t1>0)) res[R[i]][C[0]]=x;
			else                                             res[R[i]][C[0]]=u[R[i]];
		}
	}
	else if(R.size()>=2 && C.size()>=2){
		rep(i,R.size()) rep(j,C.size()) res[R[i]][C[j]]=(i+j)%2;
	}
	return {check(res,s,t,u,v),res};
}

int main(){
	int n; scanf("%d",&n);
	vector<int> s(n),t(n);
	vector<unsigned long long> u(n),v(n);
	rep(i,n) scanf("%d",&s[i]);
	rep(j,n) scanf("%d",&t[j]);
	rep(i,n) scanf("%llu",&u[i]);
	rep(j,n) scanf("%llu",&v[j]);

	unsigned long long ans[500][500]={};
	rep(b,64){
		vector<int> u2(n),v2(n);
		rep(i,n){
			u2[i]=u[i]>>b&1;
			v2[i]=v[i]>>b&1;
		}
		auto tmp=solve(n,s,t,u2,v2);
		if(!tmp.first) return puts("-1"),0;
		rep(i,n) rep(j,n) if(tmp.second[i][j]==1) ans[i][j]+=1ULL<<b;
	}
	rep(i,n) rep(j,n) printf("%llu%c",ans[i][j],j<n-1?' ':'\n');

	return 0;
}