cf287D Shifting

John Doe has found the beautiful permutation formula.

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

where k (k > 1) is an integer, the transformation parameter, r is such maximum integer that rk ≤ n. If rk = n, then elements p_rk + 1, p_rk + 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]

 

在每次变换的时候，都是取一个长度是t的区间，然后把区间的第一个元素放末尾

那么只要每次把所有这样长度为t的区间的a[kt+1]放到a[kt+t+1]即可。

比如样例的变换:

1 2 3 4 0 0 0

0 2 1 4 3 0 0

0 0 1 4 2 3 0

0 0 0 4 2 3 1

这样每次元素在数组当中的位置都会往后移一位，但是总长度还是n

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 0x7ffffff
#define pa pair<int,int>
#define mkp(a,b) make_pair(a,b)
#define pi 3.1415926535897932384626433832795028841971
using namespace std;
inline LL read()
{
    LL x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n;
int a[2000010];
int main()
{
    while (~scanf("%d",&n))
    {
        for (int i=1;i<=n;i++)a[i]=i;
        for (int k=2;k<=n;k++)
        {
            int ed=n+k-1,rp=n/k+(n%k!=0);
            for (int t=k-1+(rp-1)*k;t>=k-1;t-=k)
            {
                swap(a[ed],a[t]);
                ed=t;
            }
        }
        for (int i=n;i<=2*n-1;i++)printf("%d ",a[i]);
        puts("");
    }
}