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>
using namespace std;
template <typename T>
T next_int() {
  T x = 0, p = 1;
  char ch;
  do {
    ch = getchar();
  } while (ch <= ' ');
  if (ch == '-') {
    p = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + (ch - '0');
    ch = getchar();
  }
  return p * x;
}
string next_token() {
  char ch;
  string ans = "";
  do {
    ch = getchar();
  } while (ch <= ' ');
  while (ch > ' ') {
    ans += ch;
    ch = getchar();
  }
  return ans;
}
const long long INF = (long long)1e18 + 227 + 1;
const int INFINT = (int)1e9 + 227 + 1;
const int MAXN = (int)1e6 + 227 + 1;
const int MOD = (int)1e9 + 7;
const long double EPS = 1e-9;
long long bin_pow(long long a, long long b, long long MOD) {
  if (!b) return 1;
  long long ans = bin_pow(a, b / 2, MOD);
  ans = ans * ans % MOD;
  if (b % 2) ans = ans * a % MOD;
  return ans;
}
int main() {
  int n;
  cin >> n;
  deque<int> a;
  for (int i = 0; i < n; i++) a.push_back(i);
  for (int i = 2; i <= n; i++) {
    int len = a.size();
    int kek = len / i * i;
    if (kek >= len) kek -= i;
    for (int j = kek; j >= 0; j -= i) {
      if (j + i >= len) {
        a.push_back(a[j]);
      } else
        a[j + i] = a[j];
    }
    a.pop_front();
  }
  for (int i = 0; i < n; i++) cout << a[i] + 1 << " ";
  puts("");
}
