John Doe has found the beautiful permutation formula.

Let's take permutation p = p1, p2, ..., pn. Let's define transformation f of this permutation: 

<image>

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>
#pragma comment(linker, "/STACK:200000000")
using namespace std;
template <typename T>
inline T Abs(T x) {
  return (x >= 0) ? x : -x;
}
template <typename T>
inline T sqr(T x) {
  return x * x;
}
template <typename T>
string toStr(T x) {
  stringstream st;
  st << x;
  string s;
  st >> s;
  return s;
}
inline int nextInt() {
  int x;
  if (scanf("%d", &x) != 1) throw;
  return x;
}
inline long long nextInt64() {
  long long x;
  if (scanf("%I64d", &x) != 1) throw;
  return x;
}
inline double nextDouble() {
  double x;
  if (scanf("%lf", &x) != 1) throw;
  return x;
}
const int INF = (int)1E9;
const long long INF64 = (long long)1E18;
const long double EPS = 1E-9;
const long double PI = 3.1415926535897932384626433832795;
const int MAXN = 3001000;
int n, p[MAXN];
int main() {
  n = nextInt();
  for (int i = 0; i < (int)(n); i++) p[i] = i;
  int start = 0;
  for (int i = 2; i <= n; i++) {
    int t = -1;
    for (int x = 0; x < n; x += i) {
      int pos = x + start;
      int ppos = p[pos];
      p[pos] = t;
      t = ppos;
    }
    p[start + n] = t;
    start++;
  }
  for (int i = 0; i < (int)(n); i++) printf("%d ", p[i + start] + 1);
  return 0;
}
