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;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;

#define fi first
#define se second
#define repl(i,a,b) for(ll i=(ll)(a);i<(ll)(b);i++)
#define rep(i,n) repl(i,0,n)
#define all(x) (x).begin(),(x).end()
#define dbg(x) cout<<#x"="<<x<<endl
#define mmax(x,y) (x>y?x:y)
#define mmin(x,y) (x<y?x:y)
#define maxch(x,y) x=mmax(x,y)
#define minch(x,y) x=mmin(x,y)
#define uni(x) x.erase(unique(all(x)),x.end())
#define exist(x,y) (find(all(x),y)!=x.end())
#define bcnt __builtin_popcountll

#define INF 1e16
#define mod 1000000007

vector<ll> fib;

vector<vector<ll>> dfs(ll n){
  if(n==1){
    return vector<vector<ll>>(1,vector<ll>(1,0));
  }
  vector<vector<ll>> pre=dfs(n-1);
  vector<vector<ll>> res(n,vector<ll>(n,0));
  vector<ll> vs;
  rep(i,n-1)rep(j,n-1)res[i][j]=pre[i][j];
  rep(i,n-1)vs.push_back(i);
  ll mx=-1;
  do{
    ll sum=0;
    rep(i,n-2){
      sum+=res[vs[i]][vs[i+1]];
    }
    maxch(mx,sum);
  }while(next_permutation(all(vs)));
  mx++;

  rep(i,n-1){
    res[i][n-1]=res[n-1][i]=fib[i]*mx;
  }
  return res;
}

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

  fib.resize(10);
  fib[0]=1; fib[1]=2;
  repl(i,2,10){
    ll mn=-1;
    repl(j,fib[i-1]+1,10000){
      map<ll,bool> used;
      bool ok=true;
      rep(k,i){
        used[fib[k]]=true;
        rep(l,k)used[fib[k]+fib[l]]=true;
      }
      if(used[j])ok=false;
      used[j]=true;
      rep(k,i){
        if(used[fib[k]+j])ok=false;
        used[fib[k]+j]=true;
      }
      if(ok){
        mn=j; break;
      }
    }
    fib[i]=mn;
  }

  ll N;
  cin>>N;
  vector<vector<ll>> res=dfs(N);

  rep(i,N){
    rep(j,N){
      cout<<res[i][j]<<" ";
    }
    cout<<endl;
  }

  return 0;
}