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<algorithm>/*{{{*/
#include<cctype>
#include<cassert>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<ctime>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<vector>
using namespace std;
typedef long long lld;
typedef long double lf;
typedef unsigned long long uld;
typedef pair<int,int> pii;
#define fi first
#define se second
#define pb push_back
#define mk make_pair
#define FOR(i,a,b) for(int i=(a);i<=(b);++i)
#define ROF(i,a,b) for(int i=(a);i>=(b);--i)
namespace RA{
    int r(int p){return 1ll*rand()*rand()%p;}
    int r(int L,int R){return r(R-L+1)+L;}
}
namespace IO{
    char nc(){ static char bf[100000],*p1=bf,*p2=bf; return p1==p2&&(p2=(p1=bf)+ fread(bf,1,100000,stdin),p1==p2)?EOF:*p1++; }
    int rd(){ int res=0; char c=getchar(); while(!isdigit(c))c=getchar(); while(isdigit(c))res=res*10+c-'0',c=getchar(); return res; }
}/*}}}*/
/******************heading******************/
#define int lld
using namespace RA;
const int N=11;

int w[N][N];
int a[]={0,1,2,4,7,12,20,29,38,52,73};

int getmax(int x){
    int p[N],res=0;
    FOR(i,1,x)p[i]=i;
    do{
        int s=0;
        FOR(i,1,x-1)s+=w[p[i]][p[i+1]];
        res=max(res,s);
    }while(next_permutation(p+1,p+x+1));
    return res;
}
void make(int k){
    int m=getmax(k-1);
    m++;
    FOR(i,1,k-1)w[i][k]=w[k][i]=a[i]*m;
}
int n;
signed main(){
    scanf("%lld",&n);
    FOR(i,2,n)make(i);
    FOR(i,1,n)FOR(j,1,n)printf("%lld%c",w[i][j]," \n"[j==n]);
    return 0;
}