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;

using ll=long long;
#define int ll

#define rng(i,a,b) for(int i=int(a);i<int(b);i++)
#define rep(i,b) rng(i,0,b)
#define gnr(i,a,b) for(int i=int(b)-1;i>=a;i--)
#define per(i,b) gnr(i,0,b)
#define pb push_back
#define eb emplace_back
#define a first
#define b second
#define bg begin()
#define ed end()
#define all(x) x.bg,x.ed
#ifdef LOCAL
#define dmp(x) cerr<<__LINE__<<" "<<#x<<" "<<x<<endl
#else
#define dmp(x) void(0)
#endif

template<class t,class u> void chmax(t&a,u b){if(a<b)a=b;}
template<class t,class u> void chmin(t&a,u b){if(a>b)a=b;}

template<class t> using vc=vector<t>;
template<class t> using vvc=vc<vc<t>>;

using pi=pair<int,int>;
using vi=vc<int>;

template<class t,class u>
ostream& operator<<(ostream& os,const pair<t,u>& p){
	return os<<"{"<<p.a<<","<<p.b<<"}";
}

template<class t> ostream& operator<<(ostream& os,const vc<t>& v){
	os<<"{";
	for(auto e:v)os<<e<<",";
	return os<<"}";
}

#define mp make_pair
#define mt make_tuple

using uint=unsigned;
using ull=unsigned long long;

template<int i,class T>
void print_tuple(ostream&,const T&){
}

template<int i,class T,class H,class ...Args>
void print_tuple(ostream&os,const T&t){
	if(i)os<<",";
	os<<get<i>(t);
	print_tuple<i+1,T,Args...>(os,t);
}

template<class ...Args>
ostream& operator<<(ostream&os,const tuple<Args...>&t){
	os<<"{";
	print_tuple<0,tuple<Args...>,Args...>(os,t);
	return os<<"}";
}

void print(ll x,int suc=1){
	cout<<x;
	if(suc==1)
		cout<<endl;
	if(suc==2)
		cout<<" ";
}

ll read(){
	ll i;
	cin>>i;
	return i;
}

vi readvi(int n,int off=0){
	vi v(n);
	rep(i,n)v[i]=read()+off;
	return v;
}

template<class T>
void print(const vector<T>&v,int suc=1){
	rep(i,v.size())
		print(v[i],i==int(v.size())-1?suc:2);
}

string readString(){
	string s;
	cin>>s;
	return s;
}

template<class T>
T sq(const T& t){
	return t*t;
}

//#define CAPITAL
void yes(bool ex=true){
	#ifdef CAPITAL
	cout<<"YES"<<endl;
	#else
	cout<<"Yes"<<endl;
	#endif
	if(ex)exit(0);
}
void no(bool ex=true){
	#ifdef CAPITAL
	cout<<"NO"<<endl;
	#else
	cout<<"No"<<endl;
	#endif
	if(ex)exit(0);
}

constexpr ll ten(int n){
	return n==0?1:ten(n-1)*10;
}

const ll infLL=LLONG_MAX/3;

#ifdef int
const int inf=infLL;
#else
const int inf=INT_MAX/2-100;
#endif

//ayasii
int topbit(signed t){
	return t==0?-1:31-__builtin_clz(t);
}
int topbit(ll t){
	return t==0?-1:63-__builtin_clzll(t);
}
int popcount(signed t){
	return __builtin_popcount(t);
}
int popcount(ll t){
	return __builtin_popcountll(t);
}
bool ispow2(int i){
	return i&&(i&-i)==i;
}

bool inc(int a,int b,int c){
	return a<=b&&b<=c;
}

template<class t> void mkuni(vc<t>&v){
	sort(all(v));
	v.erase(unique(all(v)),v.ed);
}

void Check(vvc<int> g){
	int n=g.size();
	vi p(n);
	iota(all(p),0);
	vi vals;
	do{
		if(p[0]<p[n-1]){
			int a=0;
			rep(i,n-1)
				a+=g[p[i]][p[i+1]];
			vals.pb(a);
		}
	}while(next_permutation(all(p)));
	int s=vals.size();
	mkuni(vals);
	if((int)vals.size()==s){
		cerr<<vals[s-1]<<endl;
		if(vals[s-1]<=ten(11)){
			for(auto v:g)
				print(v);
			exit(0);
		}
	}
}

signed main(){
	cin.tie(nullptr);
	ios::sync_with_stdio(false);
	cout<<fixed<<setprecision(20);

	mt19937_64 rnd;

	int n;cin>>n;
	
	vvc<int> g(n,vi(n));
	g[0][1]=g[1][0]=1;
	rng(i,2,n){
		vvc<int>(i,vi(i,inf));
		
		rep(_,i){
			vi p(i+1);
			iota(all(p),0);
			vi mn(i,inf);
			int mx=0,mx2=0,mn2=inf;
			do{
				int a=-1,b=0;
				rep(j,i){
					if(j==i-2){
						if(a==-1)
							chmin(mn2,b);
					}
					if(j==i-1){
						if(a==-1)
							chmax(mx2,b);
					}
					b+=g[p[j]][p[j+1]];
					if(g[p[j]][p[j+1]]==0){
						if(a==-1)a=p[j]+p[j+1]-i;
						else{
							a=-2;
							break;
						}
					}
				}
				if(a==-2)continue;
				if(a==-1)chmax(mx,b);
				else chmin(mn[a],b);
			}while(next_permutation(all(p)));
			dmp(i);
			dmp(_);
			dmp(g);
			dmp(mx);
			dmp(mx2);
			dmp(mn);
			int j=min_element(all(mn))-mn.bg;
			g[i][j]=g[j][i]=max(mx-mn[j],mx2-mn2)+1;
		}
	}
	cerr<<g<<endl;
	Check(g);
}