Problem:
Score : 900 points

Problem Statement

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.

Sample Input 1

3

Sample Output 1

0 6 15
6 0 21
15 21 0

There are three Hamiltonian paths. The total lengths of these paths are as follows:

1 → 2 → 3: The total length is 6 + 21 = 27.

1 → 3 → 2: The total length is 15 + 21 = 36.

2 → 1 → 3: The total length is 6 + 15 = 21.

They are distinct, so the objective is met.

Sample Input 2

4

Sample Output 2

0 111 157 193
111 0 224 239
157 224 0 258
193 239 258 0

There are 12 Hamiltonian paths, with distinct total lengths.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

#define _for(i,j,N) for(int i = (j);i < (N);i++)
#define _rep(i,j,N) for(int i = (j);i <= (N);i++)
#define ALL(x) x.begin(),x.end()
#define pb push_back
#define mk make_pair

typedef long long LL;
typedef pair<int,int> Interval;

template<typename T>
ostream& operator<<(ostream& os,const vector<T>& v) {
    _for(i,0,v.size()) os << v[i] << " ";
    return os;
}

template<typename T>
ostream& operator<<(ostream& os,const set<T>& v){
    for(typename set<T>::iterator it = v.begin();it != v.end();it++)
    os << *it <<" ";
    return os;
}

template<typename T1,typename T2>
ostream& operator<<(ostream& os,const pair<T1,T2>& v){
    os << v.first <<" "<<v.second<<endl;
    return os;
}

const int maxn = 11;
LL num = 1;
LL bef = 0;
int n;
LL ans[maxn][maxn];



int main()
{
    scanf("%d",&n);
    _for(i,0,n){
        num = 1;
        bef = 0;
        _for(j,i+1,n){
            ans[i][j] = num;
            LL temp = num;
            num = temp + bef + 1;
            bef = temp;
        }
    }

     _for(i,1,n){
            num = 0;
            _for(j,0,i)
            {
                num += ans[j][n-1] + ans[j][n-2] +1;
            }
            _for(j,i+1,n){
                ans[i][j] += num;
        }
    }


    _for(i,0,n){
        for(int j = 0;j < i;j++){
                ans[i][j] = ans[j][i];
        }
    }

    _for(i,0,n){
        _for(j,0,n){
            if(j) printf(" ");
            printf("%lld",ans[i][j]);
        }
        printf("\n");
    }

    return 0;
}