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 MN = 2000010;
int N;
int a[MN];
void output(int offset) {
  for (int i = int(offset); i < int(N + offset); ++i) {
    printf("%d", a[i]);
    printf((i == N + offset - 1) ? "\n" : " ");
  }
}
int main() {
  cin >> N;
  for (int i = int(0); i < int(N); ++i) {
    a[i] = i + 1;
  }
  for (int i = int(0); i < int(N - 1); ++i) {
    int chunk = i + 2;
    int next = a[i];
    int pos = i;
    bool finish = false;
    while (!finish) {
      pos += chunk;
      if (pos >= i + N) {
        pos = i + N;
        finish = true;
      }
      swap(a[pos], next);
    }
  }
  output(N - 1);
  return 0;
}
