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 <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;
int n,s[510],t[510],l[2][510],h[2][510],now[510][510];
unsigned long long u[510],v[510],a[510][510];
bool ok=1;
inline void init()
{
	memset(l,0,sizeof l); 
	memset(h,0,sizeof h); 
	memset(now,-1,sizeof now);
}
bool solve(int k)
{
	for (int i=1;i<=n;i++)
	if (s[i]==0 && (u[i]&1))
	{
		for (int j=1;j<=n;j++)
		{
			if (now[i][j]!=-1 && now[i][j]!=1) return 0;
			if (now[i][j]==-1)
			{
				now[i][j]=1; h[1][i]++; l[1][j]++;
			}
		}
	}// and==1 all 1
	else if (s[i]==1 && !(u[i]&1))
	{
		for (int j=1;j<=n;j++)
		{
			if (now[i][j]!=-1 && now[i][j]!=0) return 0;
			if (now[i][j]==-1)
			{
				now[i][j]=0; h[0][i]++; l[0][j]++;
			}
		}
	}// or==0 all 0
	for (int j=1;j<=n;j++)
	if (t[j]==0 && (v[j]&1))
	{
		for (int i=1;i<=n;i++)
		{
			if (now[i][j]!=-1 && now[i][j]!=1) return 0;
			if (now[i][j]==-1)
			{
				now[i][j]=1; h[1][i]++; l[1][j]++;
			}
		}
	}// and==1 all 1
	else if (t[j]==1 && !(v[j]&1))
	{
		for (int i=1;i<=n;i++)
		{
			if (now[i][j]!=-1 && now[i][j]!=0) return 0;
			if (now[i][j]==-1)
			{
				now[i][j]=0; h[0][i]++; l[0][j]++;
			}
		}
	}// or==0 all 0
	for (int i=1;i<=n;i++)
	 for (int j=1;j<=n;j++)
	  if (s[i]==1 && t[j]==1 && (u[i]&1) && (v[j]&1) && now[i][j]==-1)
	  {
	  	now[i][j]=1; h[1][i]++; l[1][j]++;
	  }
	// both or 1 val=1
	//printf("lala %d %d %d %d\n",h[0][1],h[0][2],l[0][1],l[0][2]);  
	for (int i=1;i<=n;i++)
	 for (int j=1;j<=n;j++)
	  if (now[i][j]==-1)
	  {
	  	now[i][j]=0; h[0][i]++; l[0][j]++;
	  }
	// else goes 0
	 
	for (int i=1;i<=n;i++)
	if (s[i]==1 && (u[i]&1))
	{
		if (h[1][i]) continue;
		bool ff=0;
		for (int j=1;j<=n;j++)
		if (((t[j]==0 && !(v[j]&1) && l[0][j]>1)||(t[j]==1 && (v[j]&1))) && now[i][j]==0)
		{
			ff=1; now[i][j]=1; h[0][i]--; l[0][j]--; h[1][i]++; l[1][j]++;
			break;
		}
		if (!ff) return 0;
	}
	for (int j=1;j<=n;j++)
	if (t[j]==1 && (v[j]&1))
	{
		if (l[1][j]) continue;
		bool ff=0;
		for (int i=1;i<=n;i++)
		if (((s[i]==0 && !(u[i]&1) && h[0][i]>1)||(s[i]==1 && (u[i]&1))) && now[i][j]==0)
		{
			ff=1; now[i][j]=1; h[0][i]--; l[0][j]--; h[1][i]++; l[1][j]++;
			break;
		}
		if (!ff) return 0;
	}
	for (int i=1;i<=n;i++)
	{
		if (s[i]==0 && (u[i]&1) && h[1][i]!=n) return 0;
		if (s[i]==0 && !(u[i]&1) && h[0][i]==0) return 0;
		if (s[i]==1 && (u[i]&1) && h[1][i]==0) return 0;
		if (s[i]==1 && !(u[i]&1) && h[0][i]!=n) return 0;
	}
	for (int j=1;j<=n;j++)
	{
		if (t[j]==0 && (v[j]&1) && l[1][j]!=n) return 0;
		if (t[j]==0 && !(v[j]&1) && l[0][j]==0) return 0;
		if (t[j]==1 && (v[j]&1) && l[1][j]==0) return 0;
		if (t[j]==1 && !(v[j]&1) && l[0][j]!=n) return 0;
	}
	for (int i=1;i<=n;i++)
	 for (int j=1;j<=n;j++) if (now[i][j]==1) a[i][j]+=(1ull<<k);
return 1;
}
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%d",&s[i]);
	for (int i=1;i<=n;i++) scanf("%d",&t[i]);
	for (int i=1;i<=n;i++) scanf("%llu",&u[i]);
	for (int i=1;i<=n;i++) scanf("%llu",&v[i]);
	for (int i=0;i<=63 && ok;i++)
	{
		init(); ok&=solve(i);
		for (int j=1;j<=n;j++) u[j]>>=1,v[j]>>=1;
	}
	if (!ok) {printf("-1\n"); return 0;}
	for (int i=1;i<=n;i++)
	 for (int j=1;j<=n;j++) printf("%llu%c",a[i][j],(j==n?'\n':' '));
return 0;
}