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, step, start = 1;
int pos = 0;
int a[2000010];
inline void update(int Pos) {
  if (Pos >= start + n) return;
  update(Pos + step);
  a[min(Pos + step, start + n)] = a[Pos];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) a[i] = i;
  for (step = 2; step <= n; step++) {
    update(start);
    start++;
  }
  printf("%d", a[start]);
  for (int i = start + 1; i < start + n; i++) printf(" %d", a[i]);
  printf("\n");
  return 0;
}
