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=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 int n;
 int a[];
 int main()
 {
     while (~scanf("%d",&n))
     {
         for (int i=;i<=n;i++)a[i]=i;
         for (int k=;k<=n;k++)
         {
             int ed=n+k-,rp=n/k+(n%k!=);
             for (int t=k-+(rp-)*k;t>=k-;t-=k)
             {
                 swap(a[ed],a[t]);
                 ed=t;
             }
         }
         for (int i=n;i<=*n-;i++)printf("%d ",a[i]);
         puts("");
     }
 }

cf 287D