human
What is a Python code to print one way to set levels to the passages so that the objective is achieved, as follows: given the following:
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
The above is tricky. Write me a correct solution in PYTHON3.
assistant
#Here is Python code to print one way to set levels to the passages so that the objective is achieved, as follows:
.

#!usr/bin/env python3
from collections import defaultdict,deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS():return [list(x) for x in sys.stdin.readline().split()]
def S():
    res = list(sys.stdin.readline())
    if res[-1] == "\n":
        return res[:-1]
    return res
def IR(n):
    return [I() for i in range(n)]
def LIR(n):
    return [LI() for i in range(n)]
def SR(n):
    return [S() for i in range(n)]
def LSR(n):
    return [LS() for i in range(n)]

sys.setrecursionlimit(1000000)
mod = 1000000007

#A
def A():
    m,d = LI()
    ans = 0
    for i in range(1,m+1):
        for j in range(1,d+1):
            d1 = j%10
            d2 = j//10
            if d1 >= 2 and d2 >= 2 and d1*d2 == i:
                ans += 1
    print(ans)
    return

#B
def B():
    def add(i):
        while i <= m:
            bit[i] += 1
            i += i&-i

    def sum(i):
        res = 0
        while i > 0:
            res += bit[i]
            i -= i&-i
        return res

    n,k = LI()
    m = 2000
    a = LI()
    bit = [0]*(m+1)
    ans = 0
    for i in range(n):
        ans += (i-sum(a[i]))*k
        add(a[i])
    inv = pow(2,mod-2,mod)
    for i in range(n):
        ans += sum(a[i]-1)*k*(k-1)*inv%mod
    print(ans%mod)
    return

#C
def C():
    n = I()
    s = S()
    if "W" in (s[0],s[-1]):
        print(0)
        return

    return

#D
def D():
    n = I()
    m = n.bit_length()
    v = [[0]*n for i in range(n)]
    for i in range(n):
        for j in range(n):
            k = i^j
            for l in range(m):
                if k&(1<<l):
                    v[i][j] = l+1
                    break
    for i in range(n-1):
        print(*v[i][i+1:])
    return

#E
def E():

    return

#F
def F():

    return

#Solve
if __name__ == "__main__":
    D()