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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
int read()
{
    int x=0,f=1;
    char ch=getchar();
    while(ch-'0'<0||ch-'0'>9){if(ch=='-') f=-1;ch=getchar();}
    while(ch-'0'>=0&&ch-'0'<=9){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
ll re[15][15];
int dis[]={0,1,2,4,7,12,20,29,38,52,73};
int n;
int book[15];
int a[15];
int vis[15];
ll mx;
void dfs(int x,int n)
{
    if(x>n)
    {
        ll tmp=0;
        for(int i=2;i<=n;i++)  tmp=tmp+re[a[i-1]][a[i]];
        mx=max(mx,tmp);
        return;
    }
    for(int i=1;i<=n;i++)
    {
        if(!vis[i])
        {
            a[i]=x;
            vis[i]=1;
            dfs(x+1,n);
            vis[i]=0;
        }
    }
}
int main()
{
    n=read();
    re[1][2]=re[2][1]=1;
    for(int i=3;i<=n;i++)
    {
        mx=0;
        dfs(1,i-1);
        for(int j=1;j<i;j++)  re[i][j]=re[j][i]=(ll)dis[j]*(mx+1);
    }
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)  printf("%lld ",re[i][j]);
        puts("");
    }
    return 0;
}