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;
const int di[] = {-1, 0, 1, 0};
const int dj[] = {0, 1, 0, -1};
const long long MOD = 1e9 + 7;
const long long INF = 1e9;
const double EPS = 1e-5;
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  int n;
  cin >> n;
  deque<int> num(n);
  for (int i = 0; i < n; i++) num[i] = i + 1;
  for (int shft = 2; shft <= n; shft++) {
    int last = num[0];
    for (int i = shft; i < n; i += shft) swap(num[i], last);
    num.pop_front();
    num.push_back(last);
  }
  for (int i = 0; i < n; i++) cout << num[i] << " \n"[i == n - 1];
  cin.ignore(), cin.get();
}
