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 double Pi = acos(-1.0);
int A[2 * 1000002];
template <class type>
class BIT {
  type *tree;
  int maxVal;

 public:
  BIT(type *T, int n) {
    tree = T;
    maxVal = n;
    memset(tree, 0, sizeof(type) * maxVal);
  }
  void update(int id, type val) {
    while (id < maxVal) {
      tree[id] += val;
      id += id & -id;
    }
  }
  type read(int id) {
    type sum = 0;
    while (id > 0) {
      sum += tree[id];
      id -= id & -id;
    }
    return sum;
  }
  int findFirst(type val) {
    int high = maxVal - 1, low = 1;
    while (low <= high) {
      int mid((low + high) / 2);
      if (read(mid) < val)
        low = mid - 1;
      else
        high = mid + 1;
    }
    return low;
  }
  int findLast(type val) {
    int high = maxVal - 1, low = 1;
    while (low <= high) {
      int mid((low + high) / 2);
      if (read(mid) > val)
        high = mid + 1;
      else
        low = mid - 1;
    }
    return high;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  int n, c = 1;
  cin >> n;
  for (int i = 1; i <= n; i++) A[i] = i;
  for (int k = 2; k <= n; k++) {
    if (n / k * k < n) A[c + n] = A[c + n / k * k];
    for (int i = c + n / k * k; i > c; i -= k) {
      A[i] = A[i - k];
    }
    c++;
  }
  for (int i = (c); i < (c + n); i++) {
    cout << A[i] << " ";
  }
  return 0;
}
