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 <iostream>
#include <utility>
#include <set>
#include <algorithm>
#include <vector>
#define llint long long
#define mod 1000000007

using namespace std;
typedef pair<llint, llint> P;

llint n;
llint beki[15];
llint ans[15][15];

int main(void)
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n;
	beki[0] = 1;
	for(int i = 1; i <= n; i++) beki[i] = beki[i-1] * 10;
	
	for(int i = 1; i <= n; i++){
		int cnt = 0;
		for(int j = 1; j <= n; j++){
			if(i == j) continue;
			cnt++;
			ans[i][j] += beki[i-1] * cnt;
			ans[j][i] += beki[i-1] * cnt;
		}
	}
	
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= n; j++){
			cout << ans[i][j]+1 << " ";
		}
		cout << endl;
	}
	
	return 0;
}