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 syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<int,P> pip;
typedef vector<pip> vip;
const int inf=1<<30;
const ll INF=1ll<<60;
const double pi=acos(-1);
const double eps=1e-8;
const ll mod=1e9+7;
const int dx[4]={-1,0,1,0},dy[4]={0,-1,0,1};

int n;
vvl g;
vi a,b;

ll dfs(int v,ll d,int t,int N){
	if(t==N-1) return d;
	a[v]++;
	ll M=0;
	for(int u=0;u<N;u++) if(!a[u]) M=max(M,dfs(u,d+g[v][u],t+1,N));
	a[v]--;
	return M;
}

int main(){
	cin>>n;
	g=vvl(n,vl(n));
	a=b=vi(n);
	g[0][1]=g[1][0]=b[0]=1;
	set<int> st;
	st.insert(1);
	for(int i=1;i<n;i++){
		for(int j=0;j<i-1;j++) st.insert(b[j]+b[i-1]);
		int t=0;
		while(1){
			bool B=1;
			if(st.find(t)!=st.end()) B=0;
			for(int j=0;j<i;j++) if(st.find(b[j]+t)!=st.end()) B=0;
			if(B){
				b[i]=t;
				st.insert(t);
				break;
			}
			t++;
		}
	}
	for(int i=2;i<n;i++){
		ll M=0;
		for(int j=0;j<i;j++){
			a=vi(n);
			M=max(M,dfs(j,0,0,i));
		}
		for(int j=0;j<i;j++) g[i][j]=g[j][i]=(M+1)*b[j];
	}
	for(int i=0;i<n;i++) for(int j=0;j<n;j++) cout<<g[i][j]<<" \n"[j==n-1];
}