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<algorithm>
#include<cstring>
#include<cctype>
#include<cstdio>
#define rep(i,x,y) for(int i=x; i<=y; ++i)

using namespace std;
const int N=11,dis[]={0,1,2,4,7,12,20,29,38,52,73};
int n,p[N],id[N][N],tot;
typedef long long LL;
LL mx,ans[N][N];
bool vis[N];

void dfs(int x,int n)
{
	if(x>n)
	{
		LL tot=0;
		rep(i,1,n-1) tot+=ans[p[i]][p[i+1]];
		mx=max(mx,tot);
		return;
	}
	rep(i,1,n) if(!vis[i]) vis[i]=1,p[x]=i,dfs(x+1,n),vis[i]=0;
}

void solve()
{
	ans[1][2]=ans[2][1]=1;
	rep(i,3,n)
	{
		mx=0;
		dfs(1,i-1);
		rep(j,1,i-1) ans[i][j]=ans[j][i]=(mx+1)*dis[j];
	}
	rep(i,1,n)
	{
		rep(j,1,n) printf("%lld ",ans[i][j]);
		puts("");
	}
}

int main()
{
	scanf("%d",&n);
	if(n<10)
	{
		rep(i,1,n) rep(j,1,n) if(i<j) id[i][j]=tot++;
		rep(i,1,n) rep(j,1,n) if(i>j) id[i][j]=id[j][i];
		rep(i,1,n)
		{
			rep(j,1,n)
				if(i==j) printf("0 ");
				else printf("%lld ",1ll<<id[i][j]);
			puts("");
		}
		return 0;
	}
	solve();
	return 0;
}