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 size = 2000060;
const long long INF = 1000000000;
const double pi = 4 * atan(1.0);
const long long MOD = 1000000007;
int mas[size];
long long n;
int main() {
  long long m, tests;
  cin >> n;
  for (int i = 1; i <= n; ++i) mas[i - 1] = i;
  int curstart = 0;
  int curend = n;
  int dil = 2;
  while (dil != n + 1) {
    int temp;
    int prev = mas[curstart];
    for (int j = curstart + dil; j < curend; j += dil) {
      temp = mas[j];
      mas[j] = prev;
      prev = temp;
    }
    mas[curend] = prev;
    ++curend;
    ++curstart;
    dil++;
  }
  for (int i = curstart; i <= curend - 1; ++i) {
    if (i == curstart)
      printf("%d", mas[i]);
    else
      printf(" %d", mas[i]);
  }
  cout << endl;
  return 0;
}
