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>
using namespace std;
inline void fu(int &st, int &ed, vector<int> &vec, int k) {
  int tmp = vec[st];
  for (int i = st;; i += k) {
    if (i > ed + 1) i = ed + 1;
    swap(vec[i], tmp);
    if (i == ed + 1) break;
  }
  st++, ed++;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<int> vec(n * 2 + 2);
  for (int i = 1; i <= n; i++) vec[i - 1] = i;
  int st = 0, ed = n - 1;
  for (int k = 2; k <= n; k++) fu(st, ed, vec, k);
  for (int i = st; i <= ed; i++) cout << vec[i] << " \n"[i == ed];
  return 0;
}
