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 clr(x) memset(x,0,sizeof(x))
#define For(i,a,b) for (int i=(a);i<=(b);i++)
#define Fod(i,b,a) for (int i=(b);i>=(a);i--)
#define fi first
#define se second
#define kill _z_kill
#define y0 _z_y0
#define y1 _z_y1
#define x0 _z_x0
#define x1 _z_x1
#define pb(x) push_back(x)
#define mp(x,y) make_pair(x,y)
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
typedef unsigned uint;
typedef long double LD;
typedef vector <int> vi;
typedef pair <int,int> pii;
void enable_comma(){}
string tostring(char c);
string tostring(LL x);
template <class A,class B> string tostring(pair <A,B> p);
template <class A> string tostring(vector <A> v);
string tostring(char c){
	string s="";
	s+=c;
	return s;
}
string tostring(string s){
	return "\""+s+"\"";
}
string tostring(char *c){
	return tostring((string)c);
}
string tostring(LL x){
	if (x<0)
		return "-"+tostring(-x);
	if (x>9)
		return tostring(x/10)+tostring(char('0'+x%10));
	else
		return tostring(char('0'+x));
}
string tostring(int x){
	return tostring((LL)x);
}
string tostring(ull x){
	if (x>9)
		return tostring((LL)(x/10))+tostring(char('0'+x%10));
	else
		return tostring(char('0'+x));
}
string tostring(uint x){
	return tostring((LL)x);
}
string tostring(double x){
	static char res[114];
	sprintf(res,"%lf\n",x);
	return tostring(res);
}
string tostring(LD x){
	return tostring((double)x);
}
template <class A,class B> string tostring(pair <A,B> p){
	return "("+tostring(p.fi)+","+tostring(p.se)+")";
}

template <class A> string tostring(vector <A> v){
	string res="{";
	For(i,0,(int)v.size()-1){
		res+=tostring(v[i]);
		res+=i==(int)v.size()-1?"}":",";
	}
	if (res=="{")
		res+="}";
	return res;
}
template <class A> string tostring(A a,int L,int R){
	string res="{";
	For(i,L,R){
		res+=tostring(a[i]);
		res+=i==R?"}":",";
	}
	if (res=="{")
		res+="}";
	return res;
}
string tostrings(){
	return "";
}
template <typename Head,typename... Tail>
string tostrings(Head H,Tail... T){
	return tostring(H)+" "+tostrings(T...);
}
#define User_Time ((double)clock()/CLOCKS_PER_SEC)
#ifdef zzd
	#define outval(x) cerr<<#x" = "<<tostring(x)<<endl
	#define outvals(...) cerr<<"["<<#__VA_ARGS__<<"]: "<<tostrings(__VA_ARGS__)<<endl
	#define outtag(x) cerr<<"--------------"#x"---------------"<<endl
	#define outsign(x) cerr<<"<"#x">"<<endl
	#define outarr(a,L,R) cerr<<#a"["<<(L)<<".."<<(R)<<"] = "<<tostring(a,L,R)<<endl
#else
	#define outval(x) enable_comma()
	#define outvals(...) enable_comma()
	#define outtag(x) enable_comma()
	#define outsign(x) enable_comma()
	#define outarr(a,L,R) enable_comma()
#endif
#ifdef ONLINE_JUDGE
	#ifdef assert
		#undef assert
	#endif
	#define assert(x) (!(x)?\
			cout<<"Assertion failed!"<<endl<<\
			"function: "<<__FUNCTION__<<endl<<\
			"line: "<<__LINE__<<endl<<\
			"expression: "<<#x<<endl,exit(3),0:1)
#endif
LL read(){
	LL x=0,f=0;
	char ch=getchar();
	while (!isdigit(ch))
		f=ch=='-',ch=getchar();
	while (isdigit(ch))
		x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
	return f?-x:x;
}
template <class T> void ckmax(T &x,const T y){
	if (x<y)
		x=y;
}
template <class T> void ckmin(T &x,const T y){
	if (x>y)
		x=y;
}
//int Pow(int x,int y){
//	int ans=1;
//	for (;y;y>>=1,x=(LL)x*x%mod)
//		if (y&1)
//			ans=(LL)ans*x%mod;
//	return ans;
//}
//void Add(int &x,int y){
//	if ((x+=y)>=mod)
//		x-=mod;
//}
//void Del(int &x,int y){
//	if ((x-=y)<0)
//		x+=mod;
//}
//int Add(int x){
//	return x>=mod?x-mod:x;
//}
//int Del(int x){
//	return x<0?x+mod:x;
//}
//int md(LL x){
//	return (x%mod+mod)%mod;
//}
const int N=233;
int n;
LL g[N][N];
namespace test{
	void main(){
		unordered_map <LL,int> vis;
		vi id(n);
		iota(id.begin(),id.end(),1);
		do {
			LL now=0;
			For(i,1,(int)id.size()-1)
				now+=g[id[i-1]][id[i]];
			assert(now<=1e11);
			if ((++vis[now])>2)
				assert(0);
		} while (next_permutation(id.begin(),id.end()));
	}
}
int main(){
	n=read();
	g[n-1][n]=1;
	Fod(i,n-2,1){
		For(j,i+2,min(n,i+3))
			g[i][n]+=g[i+1][j];
		if (i==n-2)
			g[i][n]++;
		g[i][n-1]=g[i][n]*2-1;
		if (i<n-2)
			g[i][n-1]--;
		if (i<n-3)
			g[i][n-1]--;
		Fod(j,n-2,i+1)
			g[i][j]=g[i][j+1]+g[i][j+2]+g[i][n]-2;
	}
	For(i,1,n)
		For(j,i+1,n)
			g[j][i]=g[i][j];
	For(i,1,n){
		For(j,1,n)
			printf("%lld ",g[i][j]);
		puts("");
	}
	#ifdef zzd
		test::main();
	#endif
	return 0;
}