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;

set<int> s;

vector<int> v,e;
long long int g[10][10];
int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	v.emplace_back(1);
	v.emplace_back(2);
	int now=3;
	s.insert(1);
	s.insert(2);
	s.insert(3);
	for(int i=2;i<=8;i++){
		while(true){
			bool sw=0;
			for(int j=0;j<i;j++)
				if(s.find(v[j]+now)!=s.end())
					sw=true;
			if(s.find(now)!=s.end())	sw=true;
			if(!sw){
				for(int j=0;j<i;j++)
					s.insert(v[j]+now);
				v.emplace_back(now);
				now++;
				break;
			}
			now++;
		}
	}
	g[0][1]=g[1][0]=1;
	e.emplace_back(0);
	e.emplace_back(1);
	long long int ans=0;
	for(int i=2;i<10;i++){
		long long int M=0;
		do{
			long long int tmp=0;
			for(int j=0;j<i-1;j++)
				tmp+=g[e[j]][e[j+1]];
			M=max(M,tmp);
		}while(next_permutation(e.begin(),e.end()));
		M++;
		e.emplace_back(i);
		sort(e.begin(),e.end());
		for(int j=0;j<i;j++)
			g[j][i]=g[i][j]=M*v[j];
	}
	int n;
	cin>>n;
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++)
			cout<<g[i][j]<<' ';
		cout<<endl;
	}
}