John Doe has found the beautiful permutation formula.

Let's take permutation *p* = *p*<sub class="lower-index">1</sub>, *p*<sub class="lower-index">2</sub>, ..., *p*<sub class="lower-index">*n*</sub>. Let's define transformation *f* of this permutation: 

where *k* (*k* &gt; 1) is an integer, the transformation parameter, *r* is such maximum integer that *rk* ≤ *n*. If *rk* = *n*, then elements *p*<sub class="lower-index">*rk* + 1</sub>, *p*<sub class="lower-index">*rk* + 2</sub> and so on are omitted. In other words, the described transformation of permutation *p* cyclically shifts to the left each consecutive block of length *k* and the last block with the length equal to the remainder after dividing *n* by *k*. 

John Doe thinks that permutation *f*(*f*( ... *f*(*p* = [1, 2, ..., *n*], 2) ... , *n* - 1), *n*) is beautiful. Unfortunately, he cannot quickly find the beautiful permutation he's interested in. That's why he asked you to help him.

Your task is to find a beautiful permutation for the given *n*. For clarifications, see the notes to the third sample.
A single line contains integer *n* (2 ≤ *n* ≤ 10<sup class="upper-index">6</sup>).
n = int(input())

a = [0] * (n + 1)

for i in range(1, n + 1):
    a[i] = i

for i in range(2, n + 1):
    tmp = a[i - 1]
    j = i
    for c in range(0, n // i):
        if (i - 1 + j) % n == 0:
            tmp, a[n] = a[n], tmp
        else:
            tmp, a[(i - 1 + j) % n] = a[(i - 1 + j) % n], tmp
    if n % i != 0:
        a[i - 1] = tmp

print(a[n], end=' ')
for i in range(1, n):
    print(a[i], end=' ')
