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;
int n;
void print(int *p, int sh) {
  for (int _n(n), i(0); i < _n; i++) {
    cout << p[sh] + 1 << " ";
    sh++;
    if (sh == n) sh = 0;
  }
  cout << endl;
}
void shift(int *a, int &sh, int k) {
  int t = a[sh];
  for (int i = 0; i < n;) {
    i += k;
    if (i > n) i = n;
    int idx = (i + sh) % n;
    swap(t, a[idx]);
  }
  sh++;
  if (sh == n) sh = 0;
}
int p[1000100];
int main() {
  cin >> n;
  for (int _n(n), i(0); i < _n; i++) p[i] = i;
  int sh = 0;
  for (int _n(n + 1), k(2); k < _n; k++) shift(p, sh, k);
  print(p, sh);
  return 0;
}
