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<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<map>
#define ll long long
#define mod 998244353
#define maxn 200010
inline ll read()
{
	ll x=0; char c=getchar(),f=1;
	for(;c<'0'||'9'<c;c=getchar())if(c=='-')f=-1;
	for(;'0'<=c&&c<='9';c=getchar())x=x*10+c-'0';
	return x*f;
}
inline void write(ll x)
{
	static int buf[20],len; len=0;
	if(x<0)x=-x,putchar('-');
	for(;x;x/=10)buf[len++]=x%10;
	if(!len)putchar('0');
	else while(len)putchar(buf[--len]+'0');
}
inline void writeln(ll x){write(x); putchar('\n');}
inline void writesp(ll x){write(x); putchar(' ');}
ll cnt[30],base[30],now[30];
int a[30],mx[30],v[1010];
ll mp[15][15];
int n;
std::map<ll,int>mark;
int vis[maxn];
int dfs(int S,int now,int dep,int n,ll dis)
{
	if(dep==n&&now>S){
		if(mark[dis])return 0;
		mark[dis]=1;
		return 1;
	}
	for(int i=1;i<=n;i++)
		if(!vis[i]){
			vis[i]=1;
			if(!dfs(S,i,dep+1,n,dis+mp[now][i]))return 0;
			vis[i]=0;
		}
	return 1;
}
ll calc(int n)
{
	mark.clear();
	mark[0]=1;
	memset(vis,0,sizeof(vis));
	for(int i=1;i<=n;i++){
		vis[i]=1;
		dfs(i,i,1,n,0);
		vis[i]=0;
	}
	return (*(--mark.end())).first;
}
int check()
{
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			if(i==j&&mp[i][j])return 0;
			else if(i!=j&&(mp[i][j]<1||mp[i][j]>1e11))return 0;
	int flag=1;
	mark.clear();
	memset(vis,0,sizeof(vis));
	for(int i=1;i<=n;i++){
		vis[i]=1;
		flag&=dfs(i,i,1,n,0);
		vis[i]=0;
	}
	return flag&&calc(n)<=1e11;
}
int main()
{
	n=read();
	int k=1;
	a[0]=0;
	for(int i=1;i<=n;i++){
		while(1){
			int flag=1;
			for(int j=0;j<i;j++)
				if(v[k+a[j]])flag=0;
			if(flag)break;
			++k;
		}
		a[i]=k;
		for(int j=0;j<i;j++)
			v[a[i]+a[j]]=1;
	}
//	 for(int i=1;i<=n;i++)
//	 	printf("%d %d ***\n",i,a[i]);
	for(int i=2;i<=n;i++){
		ll mx=calc(i-1)+1;
		for(int j=1;j<i;j++)
			mp[i][j]=mp[j][i]=mx*a[j];
	}
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++)
			writesp(mp[i][j]);
		putchar('\n');
	}
//	puts(check()?"AC":"WA");
	return 0;
}