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;
void test(int n) {
  vector<int> a(n + 1);
  for (int i = 1; i <= n; i++) a[i] = i;
  for (int k = 2; k <= n; k++) {
    for (int i = 1; i < n; i++)
      if (i % k != 0) {
        swap(a[i], a[i + 1]);
      }
  }
  printf(":: %d\n", n);
  for (int i = 1; i <= n; i++) printf("%d ", a[i]);
  puts("");
}
const int MAXN = 1000006;
int n, a[MAXN * 2];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) a[i] = i;
  int first = 1;
  for (int k = 2; k <= n; k++, first++) {
    int tmp = a[first], tmp2;
    for (int i = first; i < first + n; i += k) {
      if (i + k < first + n) {
        tmp2 = a[i + k];
        a[i + k] = tmp;
        tmp = tmp2;
      } else {
        a[first + n] = tmp;
      }
    }
  }
  for (int i = first; i < first + n; i++) {
    printf("%d%c", a[i], i == first + n - 1 ? '\n' : ' ');
  }
}
