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 inf = ~0U >> 1;
const long long INF = ~0ULL >> 1;
const double pi = acos(-1.0);
template <class T>
inline T sqr(T a) {
  return a * a;
}
template <class T>
inline T min(T a, T b, T c) {
  return min(min(a, b), c);
}
template <class T>
inline T max(T a, T b, T c) {
  return max(max(a, b), c);
}
template <class T>
inline void read(T &n) {
  char c;
  for (c = getchar(); !(c >= '0' && c <= '9'); c = getchar())
    ;
  n = c - '0';
  for (c = getchar(); c >= '0' && c <= '9'; c = getchar()) n = n * 10 + c - '0';
}
int pw(int base, int n, int mo) {
  if (n == 0) return 1;
  if (n == 1) return base;
  int tmp = pw(base, n >> 1, mo);
  tmp = (long long)tmp * tmp % mo;
  if (n & 1) tmp = (long long)tmp * base % mo;
  return tmp;
}
int N, K, a[3000000];
int main() {
  scanf("%d", &N);
  int fst = 1;
  for (int i = (1); i <= (N); ++i) a[i] = i;
  for (int k = (2); k <= (N); ++k) {
    int i;
    for (i = fst; i + k <= fst + N - 1; i += k)
      ;
    for (int j = i; j >= fst; j -= k) swap(a[j], a[min(j + k, fst + N)]);
    fst++;
  }
  for (int i = (fst); i <= (fst + N - 1); ++i) printf("%d ", a[i]);
  return 0;
}
