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>
int a[2000005];
int n, i, j;
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    a[i] = i;
  }
  for (i = 2; i <= n; i++) {
    int b = i - 1, g;
    if (n % i == 0) {
      g = n / i;
      for (j = g; j >= 1; j--) {
        a[b + j * i] = a[b + j * i - i];
      }
    } else {
      g = (n - (n % i)) / i;
      a[b + n] = a[g * i + b];
      for (j = g; j >= 1; j--) {
        a[b + j * i] = a[b + j * i - i];
      }
    }
  }
  for (i = 1; i <= n; i++) {
    printf("%d%c", a[i + n - 1], i == n ? '\n' : ' ');
  }
}
