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 <cstdlib>
#include <algorithm>
#include <vector>
#include <cstring>
#include <queue>
#include <set>
#include <map>
#include <functional>
#include <cmath>
#include <cassert>
#include <string>
#define SIZE 12
#define MX 205

using namespace std;
typedef long long int ll;
typedef pair <int,int> P;

ll ans[SIZE][SIZE];
ll fib[SIZE];
bool use[MX];

void make()
{
	fib[0]=1;
	fib[1]=2;
	for(int i=2;i<SIZE;i++)
	{
		for(int j=0;j<i;j++)
		{
			use[fib[j]]=true;
			for(int k=j+1;k<i;k++) use[fib[j]+fib[k]]=true;
		}
		fib[i]=1;
		while(1)
		{
			bool up=!use[fib[i]];
			for(int j=0;j<i;j++) if(use[fib[j]+fib[i]]) up=false;
			if(up) break;
			fib[i]++;
		}
	}
	ll mul=1;
	ans[0][0]=0;
	for(int i=1;i<SIZE;i++)
	{
		ans[i][i]=0;
		for(int j=0;j<i;j++) ans[i][j]=ans[j][i]=fib[j]*mul;
		if(i==1) mul*=2LL;
		else
		{
			mul*=fib[i-1]+fib[i-2];
			mul++;
		}
	}
}
int main()
{
	make();
	int n;
	scanf("%d",&n);
	ll mx=0;
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<n;j++)
		{
			printf("%lld ",ans[i][j]);
			mx=max(mx,ans[i][j]);
		}puts("");
	}
	/*
	vector <ll> vec;
	vector <int> ord;
	for(int i=0;i<n;i++) ord.push_back(i);
	do
	{
		ll sum=0;
		for(int i=0;i+1<n;i++) sum+=ans[ord[i]][ord[i+1]];
		vec.push_back(sum);
	}while(next_permutation(ord.begin(),ord.end()));
	sort(vec.begin(),vec.end());
	vector <ll> vx=vec;
	vx.erase(unique(vx.begin(),vx.end()),vx.end());
	printf("%d %d : %lld\n",vx.size(),vec.size(),vec[vec.size()-1]);
	//printf("%lld\n",mx);*/
	return 0;
}