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 Python (3.4.3) (judge verdict: Accepted)

def solve(n):
    if n == 2:
        print('0 1\n1 0')
        return

    can = []
    ban = set()

    k = 1
    while len(can) < 9:
        if k in ban or any(k + l in ban for l in can):
            k += 1
            continue
        ban.add(k)
        ban.update(k + l for l in can)
        can.append(k)
        k += 1

    ans = [[0] * n for _ in [0] * n]
    ans[0][1] = ans[1][0] = 1
    longest_path = 1
    for i in range(2, n):
        np = longest_path + 1
        for j, c in zip(range(i), can):
            ans[i][j] = ans[j][i] = np * c
        longest_path += ans[i][i - 1] + ans[i][i - 2] - ans[i - 1][i - 2]

    print('\n'.join(' '.join(map(str, row)) for row in ans))


n = int(input())
solve(n)