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;
struct Foo {
  int n, k, p;
  Foo() { this->n = 0; }
  Foo(int n, int p, int k) {
    this->n = n;
    this->k = k;
    this->p = p;
  }
};
Foo foo[2 << 20];
int answer[1 << 20];
int main() {
  cin >> n;
  for (int p = 0; p < n; p++) {
    foo[p + 2] = Foo(p + 1, p, 2);
  }
  for (int k = 2; k <= n; k++) {
    for (int i = (2 * n - 1) / k * k; i > 0; i -= k) {
      if (foo[i].n == 0) continue;
      int newPosition = n < i ? n - 1 : i - 1;
      int newK = k + 1;
      foo[i].p = newPosition;
      foo[i].k = newK;
      foo[newPosition + newK] = foo[i];
      foo[i].n = 0;
    }
  }
  for (int i = 0; i < 2 << 20; i++) {
    if (foo[i].n == 0) continue;
    answer[foo[i].p - (n + 1 - foo[i].k)] = foo[i].n;
  }
  for (int i = 0; i < n; i++) {
    printf("%d ", answer[i]);
  }
  return 0;
}
