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;
#define ll long long
#define ui unsigned int
#define ull unsigned long long
#define db long double
#define pii pair<int,int>
#define pli pair<ll,int>
#define X first
#define Y second
#define mp make_pair
#define pb push_back
#define vi vector<int>
#define vii vector<vi>
#define lb lower_bound
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,b,a) for(int i=(b);i>=(a);--i)
#define rep0(i,a,b) for(int i=(a);i<(b);++i)
#define fore(i,a) for(int i=0;i<a.size();++i)
#define gc() getchar()
#define ls x<<1,l,m
#define rs x<<1|1,m+1,r
inline ll rd()
{
	ll x=0,w=1;char c=gc();while(!isdigit(c)&&c!='-')c=gc();
	if(c=='-')c=gc(),w=-1;while(isdigit(c))x=x*10+c-48,c=gc();return x*w;
}
const int a[11]={0,1,2,4,7,12,20,29,38,52,73};
int n,p[11];ll m,e[11][11];
int main()
{
	n=rd();m=0;
	rep(i,2,n)
	{
		rep(j,1,i-1)e[i][j]=e[j][i]=(m+1)*a[j];
		if(i==n)break;
		m=0;rep(j,1,i)p[j]=j;
		do
		{
			ll s=0;rep(j,1,i-1)s+=e[p[j]][p[j+1]];m=max(m,s);
		}while(next_permutation(p+1,p+i+1));
	}
	rep(i,1,n)
	{
		rep(j,1,n)printf("%lld ",e[i][j]);
		puts("");
	}
	return 0;
}