AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.

For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:

* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.



Your task is to set levels to the passages so that the highest level of a passage is minimized.

Constraints

* N is an integer between 2 and 500 (inclusive).

Input

Input is given from Standard Input in the following format:


N


Output

Print one way to set levels to the passages so that the objective is achieved, as follows:


a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}


Here a_{i,j} is the level of the passage connecting Room i and Room j.

If there are multiple solutions, any of them will be accepted.

Example

Input

3


Output

1 2
1
def ans(n):
  if n==1:
    return [['0','1'],['0','0']]
  else:
    l=[['0' for i in range(2**n)] for j in range(2**n)]
    ans2=ans(n-1)
    for i in range(2**n):
      for j in range(2**n):
        if i<2**(n-1) and j<2**(n-1):
          l[i][j]=ans2[i][j]
        elif i<2**(n-1) and j>=2**(n-1):
          l[i][j]=str(n)
        elif i>=2**(n-1) and j<2**(n-1):
          l[i][j]='0'
        else:
          l[i][j]=ans2[i-2**(n-1)][j-2**(n-1)]
    return l
import math
n=int(input())
N=math.ceil(math.log(n,2))
ans=ans(N)
for i in range(n-1):
  print(' '.join(ans[i][i+1:n]))