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>
#define MOD 200000007LL
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int n;
ll twop[55];
ll ans[51][51];
int ord[15];

int main(void){
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		ord[i]=i;
	}
	int sz=n*(n-1)/2;
	std::random_device rnd;
    std::mt19937 mt(rnd());
	vector<ll> vec;
	map<ll,int> cnt;
	while(1){
		int pos=0;
		vec.clear();
		cnt.clear();
		ll p=1;
		for(int i=0;i<20;i++){
			vec.push_back(p);
			p*=2LL;
		}
		while(vec.size()<sz){
			ll val=mt()%MOD;
	    	vec.push_back(val);
	    }
	    for(int i=0;i<n;i++){
			for(int j=i+1;j<n;j++){
				ans[i][j]=vec[pos];
				ans[j][i]=vec[pos];
				pos++;
			}
		}
		bool flag=true;
		do{
			ll sum=0;
			for(int i=0;i<n-1;i++){
				sum+=ans[ord[i+1]][ord[i]];
			}
			cnt[sum]++;
			if(cnt[sum]>=3){
				flag=false;
				break;
			}
		}while(next_permutation(ord,ord+n));
		if(flag)break;
	}
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			printf("%lld%c",ans[i][j],(j+1)==n?'\n':' ');
		}
	}
	return 0;
}