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;
template <class A, class B>
ostream &operator<<(ostream &out, const pair<A, B> &a) {
  return out << "(" << a.first << "," << a.second << ")";
}
template <class A>
ostream &operator<<(ostream &out, const vector<A> &a) {
  for (const A &it : a) out << it << " ";
  return out;
}
template <class A, class B>
istream &operator>>(istream &in, pair<A, B> &a) {
  return in >> a.first >> a.second;
}
template <class A>
istream &operator>>(istream &in, vector<A> &a) {
  for (A &i : a) in >> i;
  return in;
}
ifstream cinn("in.txt");
ofstream coutt("out.txt");
long long poww(const long long &a, long long b,
               const long long &m = 1000000007) {
  if (b == 0) return 1;
  long long first = poww(a, b / 2, m);
  first = first * first % m;
  if (b & 1) first = first * a % m;
  return first;
}
long long gcd(long long a, long long b) {
  if (a > b) swap(a, b);
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long rint(long long l, long long r) { return rand() % (r - l + 1) + l; }
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long i, j, first, n, m;
  cin >> n;
  vector<long long> arr(2 * n + 5);
  for (i = 1; i < n + 1; i++) {
    arr[i] = i;
  }
  long long st = 1;
  for (i = 2; i < n + 1; i++) {
    long long cur = arr[st];
    long long pos = st;
    while (pos - st + 1 < n) {
      long long nxt = pos + i;
      if (nxt - st >= n) break;
      long long temp = arr[nxt];
      arr[nxt] = cur;
      pos = nxt;
      cur = temp;
    }
    arr[st + n] = cur;
    st++;
  }
  for (i = 0; i < n; i++) {
    cout << arr[st + i] << " ";
  }
  return 0;
}
