Problem: Diverta City is a new city consisting of N towns numbered 1, 2, ..., N.

The mayor Ringo is planning to connect every pair of two different towns with a bidirectional road. The length of each road is undecided.

A Hamiltonian path is a path that starts at one of the towns and visits each of the other towns exactly once. The reversal of a Hamiltonian path is considered the same as the original Hamiltonian path.

There are N! / 2 Hamiltonian paths. Ringo wants all these paths to have distinct total lengths (the sum of the lengths of the roads on a path), to make the city diverse.

Find one such set of the lengths of the roads, under the following conditions:

* The length of each road must be a positive integer.
* The maximum total length of a Hamiltonian path must be at most 10^{11}.

Constraints

* N is a integer between 2 and 10 (inclusive).

Input

Input is given from Standard Input in the following format:


N


Output

Print a set of the lengths of the roads that meets the objective, in the following format:


w_{1, 1} \ w_{1, 2} \ w_{1, 3} \ ... \ w_{1, N}
w_{2, 1} \ w_{2, 2} \ w_{2, 3} \ ... \ w_{2, N}
:  :     :
w_{N, 1} \ w_{N, 2} \ w_{N, 3} \ ... \ w_{N, N}


where w_{i, j} is the length of the road connecting Town i and Town j, which must satisfy the following conditions:

* w_{i, i} = 0
* w_{i, j} = w_{j, i} \ (i \neq j)
* 1 \leq w_{i, j} \leq 10^{11} \ (i \neq j)



If there are multiple sets of lengths of the roads that meet the objective, any of them will be accepted.

Examples

Input

3


Output

0 6 15
6 0 21
15 21 0


Input

4


Output

0 111 157 193
111 0 224 239
157 224 0 258
193 239 258 0

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define LL long long
#define db double

using namespace std;
const int N=500+10,M=100+10;
int rd()
{
    int x=0,w=1;char ch=0;
    while(ch<'0'||ch>'9'){if(ch=='-') w=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+(ch^48);ch=getchar();}
    return x*w;
}
LL f[12],sq[12],an[12][12],nx;
int n,vs[233];
bool ck(int j,int i)
{
	for(int k=0;k<i;++k) if(vs[sq[k]+j]) return 0;
	return 1;
}
int p[12],v[12],cn;
void dfs(int o,LL ns)
{
	if(o>n){nx=max(nx,ns);return;}
	for(int i=1;i<=n;++i)
		if(!v[i])
		{
			v[i]=1,p[o]=i;
			dfs(o+1,ns+an[p[o-1]][p[o]]);
			v[i]=0;
		}
}

int main()
{
	n=rd();
	sq[1]=1,vs[1]=1;
	for(int i=2,j=2;i<=n;++i,++j)
	{
		while(!ck(j,i)) ++j;
		sq[i]=j;
		for(int k=0;k<i;++k) vs[sq[k]+sq[i]]=1;
	}
	memcpy(f,sq,sizeof(sq));
	for(int i=2;i<=n;++i)
	{
		for(int j=1;j<i;++j) an[i][j]=an[j][i]=sq[j];
		nx=0;
		for(int j=1;j<=n;++j) v[j]=1,p[1]=j,dfs(2,0),v[j]=0;
		for(int j=1;j<=n;++j) sq[j]=f[j]*(nx+1);
	}
	for(int i=1;i<=n;++i)
	{
		for(int j=1;j<=n;++j) printf("%lld ",an[i][j]);
		puts("");
	}
	return 0;
}