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>
using namespace std;

typedef long long lint;
typedef long double louble;

#define so(a) ((int)((a).size()))

template<typename T1,typename T2> inline T1 max(T1 a,T2 b){return a<b?b:a;}
template<typename T1,typename T2> inline T1 min(T1 a,T2 b){return a<b?a:b;}

const char lf = '\n';

namespace ae86
{
	const int bufl = 1 << 15;

	char buf[bufl],*s=buf,*t=buf;

	inline int fetch()
	{
		if(s==t){t=(s=buf)+fread(buf,1,bufl,stdin);if(s==t)return EOF;}
		return *s++;
	}

	inline int ty()
	{
		int a=0;int b=1,c=fetch();
		while(!isdigit(c))b^=c=='-',c=fetch();
		while(isdigit(c))a=a*10+c-48,c=fetch();
		return b?a:-a;
	}
}
using ae86::ty;

const int _ = 11;
const int biao [] = { 0,1,2,4,7,12,20,29,38,52,73 };

int n;
lint dis[_][_]={0};

int main()
{
	ios::sync_with_stdio(0),cout.tie(nullptr);

	n=ty();

	lint now=0;
	for(int i=2;i<=n;i++)
	{
		for(int j=1;j<i;j++)
			dis[i][j]=dis[j][i]=(now+1)*biao[j];
		now=now-dis[i-1][i-2]+dis[i][i-1]+dis[i][i-2];
	}

	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			cout<<dis[i][j]<<" \n"[j==n];

	return 0;
}