John Doe has found the beautiful permutation formula.

Let's take permutation p = p1, p2, ..., pn. Let's define transformation f of this permutation: 

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where k (k > 1) is an integer, the transformation parameter, r is such maximum integer that rk ≤ n. If rk = n, then elements prk + 1, prk + 2 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.

Input

A single line contains integer n (2 ≤ n ≤ 106).

Output

Print n distinct space-separated integers from 1 to n — a beautiful permutation of size n.

Examples

Input

2


Output

2 1 


Input

3


Output

1 3 2 


Input

4


Output

4 2 3 1 

Note

A note to the third test sample: 

  * f([1, 2, 3, 4], 2) = [2, 1, 4, 3]
  * f([2, 1, 4, 3], 3) = [1, 4, 2, 3]
  * f([1, 4, 2, 3], 4) = [4, 2, 3, 1]
#include <bits/stdc++.h>
int n;
int p[2 * 1000050];
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n); i++) p[i] = i;
  int head = 1, end = n;
  int ptr, tmp_1, tmp_2;
  for (int i = (2); i <= (n); i++) {
    ptr = head;
    tmp_1 = p[head];
    while (ptr + i <= end) {
      tmp_2 = p[ptr + i];
      p[ptr + i] = tmp_1;
      tmp_1 = tmp_2;
      ptr += i;
    }
    p[end + 1] = tmp_1;
    head++;
    end++;
  }
  for (int i = (head); i <= (end); i++) printf("%d ", p[i]);
  return 0;
}
