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 PYTHON3):

Solution:
from itertools import combinations, permutations

N = int(input())

# 整数列の生成
# s = [1]
# while len(s) < 10 :
    # i = s[-1] + 1
    
    # while True :
        # path = s.copy() + [i]
        # flag = True
        # for comb in combinations(s + [i], 2) :
            # if not sum(comb) in path :
                # path.append(sum(comb))
            # else :
                # flag = False
                # break
        # if flag :
            # s.append(i)
            # break
        # else :
            # i += 1
            
s = [1, 2, 4, 7, 12, 20, 29, 38, 52, 73]

w = [[0] * 10 for _ in range(10)]
w[0][1] = w[1][0] = 1

for n in range(3, N + 1) :
    # 最長経路探索
    M = 0
    for perm in permutations(range(n-1), n-1) :
        tmp = 0
        for i in range(n-2) :
            tmp += w[perm[i]][perm[i+1]]
        M = max(M, tmp)
    
    M += 1
    
    # 新規割当
    for i in range(n-1) :
        w[i][n-1] = w[n-1][i] = M * s[i]
        
        
for i in range(N) :
    print(' '.join([str(j) for j in w[i][:N]]))