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>
char Gc() {
  static char now[1 << 20], *S, *T;
  if (T == S) {
    T = (S = now) + std::fread(now, 1, 1 << 20, stdin);
    if (T == S) return EOF;
  }
  return *S++;
}
template <typename T>
void Read(T &x) {
  x = 0;
  int f = 1;
  char c;
  while ((c = Gc()) < '0' || c > '9')
    if (c == '-') f = -1;
  x = c - '0';
  while ((c = Gc()) >= '0' && c <= '9') x = x * 10 + c - '0';
  x *= f;
}
template <typename T, typename... Args>
void Read(T &x, Args &...args) {
  Read(x);
  Read(args...);
}
template <typename T>
void checkmax(T &x, T y) {
  if (x < y) x = y;
}
template <typename T>
void checkmin(T &x, T y) {
  if (x > y) x = y;
}
int n, a[2000001];
int main(int argc, char const *argv[]) {
  Read(n);
  std::iota(a + 1, a + n + 1, 1);
  int t;
  for (int i = 2; i <= n; i++) {
    t = 0;
    for (int j = i - 1; j < i + n - 1; j += i) {
      std::swap(t, a[j]);
    }
    a[i + n - 1] = t;
  }
  for (int i = n; i < 2 * n; i++) std::printf("%d ", a[i]);
  return 0;
}
