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>
typedef long long ll;
ll gi(){
	ll x=0,f=1;
	char ch=getchar();
	while(!isdigit(ch))f^=ch=='-',ch=getchar();
	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
	return f?x:-x;
}
std::mt19937 rnd(time(NULL));
#define rand rnd
#define pr std::pair<int,int>
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
template<class T>void cxk(T&a,T b){a=a>b?a:b;}
template<class T>void cnk(T&a,T b){a=a<b?a:b;}
#ifdef mod
int pow(int x,int y){
	int ret=1;
	while(y){
		if(y&1)ret=1ll*ret*x%mod;
		x=1ll*x*x%mod;y>>=1;
	}
	return ret;
}
template<class Ta,class Tb>void inc(Ta&a,Tb b){a=a+b>=mod?a+b-mod:a+b;}
template<class Ta,class Tb>void dec(Ta&a,Tb b){a=a>=b?a-b:a+mod-b;}
#endif
int shit[]={0,1,2,4,7,12,20,29,38,52,73};
ll w[11][11];
int p[11];
int main(){
#ifdef LOCAL
	freopen("in.in","r",stdin);
	//freopen("out.out","w",stdout);
#endif
	int n=gi();
	w[1][2]=w[2][1]=1;ll M=1;
	for(int x=3;x<=n;++x){//insert node x
		for(int i=1;i<x;++i)w[x][i]=w[i][x]=shit[i]*(M+1);
		for(int i=1;i<=x;++i)p[i]=i;
		do{
			if(p[1]>p[x])continue;
			ll sum=0;
			for(int i=1;i<x;++i)sum+=w[p[i]][p[i+1]];
			cxk(M,sum);
		}while(std::next_permutation(p+1,p+x+1));
	}
	assert(M<=1e11);
	for(int i=1;i<=n;++i){
		for(int j=1;j<=n;++j)printf("%lld ",w[i][j]);
		puts("");
	}
	return 0;
}