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>
#define IL inline
#define _ 105
#define ll long long
using namespace std;

IL int gi(){
	int data=0,fu=1; char ch=0; while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
	if(ch=='-')fu=-1,ch=getchar();
	while('0'<=ch&&ch<='9')data=(data<<1)+(data<<3)+(ch^48),ch=getchar(); return data*fu;
}

int n,q[_],p[_]; bool Ban[1000005]; ll E[_][_];
map<ll,int>M;

IL ll Halmilton(int nc){
	ll ret=0,len=0;
	for(int i=1;i<=nc;++i)p[i]=i;
	do{
		len=0;
		for(int i=2;i<=nc;++i)len+=E[p[i-1]][p[i]];
		ret=max(ret,len);
	}while(next_permutation(p+1,p+nc+1)); return ret;
}

int main(){
	n=gi();
	int l=1,nq=0; bool flag;
	while(nq<n){
		flag=true;
		for(int j=1;j<=nq;++j)if(Ban[l+q[j]])flag=false;
		if(Ban[l])flag=false;
		if(flag){
			for(int j=1;j<=nq;++j)Ban[l+q[j]]=true;
			q[++nq]=l;
		}
		++l;
	}
	E[1][2]=E[2][1]=q[1];
	for(int i=3;i<=n;++i){
		ll m=Halmilton(i-1);
		for(int j=1;j<i;++j)E[i][j]=E[j][i]=1ll*(m+1)*q[j];
	}
	for(int i=1;i<=n;++i,puts(""))
		for(int j=1;j<=n;++j)cout<<E[i][j]<<" ";
	return 0;
}