Description

Let‘s call the following process a transformation of a sequence of length $\mathrm{nn\; .}$

If the sequence is empty, the process ends. Otherwise, append the greatest common divisor (GCD) of all the elements of the sequence to the result and remove one arbitrary element from the sequence. Thus, when the process ends, we have a sequence of

You are given an integer sequence $\mathrm{1,2,\dots ,n1,2,\dots ,n\; .\; Find\; the\; lexicographically\; maximum\; result\; of\; its\; transformation.}$

A sequence ${\mathrm{a1,a2,\dots ,ana1,a2,\dots ,an\; is\; lexicographically\; larger\; than\; a\; sequence\; b1,b2,\dots ,bnb1,b2,\dots ,bn\; ,\; if\; there\; is\; an\; index}}^{}$

Input

The first and only line of input contains one integer $\mathrm{nn\; (1\le n\le 1061\le n\le 106\; ).}$

Output

Output $\mathrm{nn\; integers\; —\; the\; lexicographically\; maximum\; result\; of\; the\; transformation.}$

Sample Input

3

1 1 3 

2

1 2 

1

1 

Sample Output

Hint

In the first sample the answer may be achieved this way:

• Append GCD$(1,2,3)=1(1,2,3)=1\; ,\; remove\; 22\; .$
• Append GCD$(1,3)=1(1,3)=1\; ,\; remove\; 11\; .$
• Append GCD$(3)=3(3)=3\; ,\; remove\; 33\; .$

We get the sequence $[1,1,3][1,1,3]\; as\; the\; result.$

盡可能的讓大的gcd值盡快出現。

有一條規則可以推出來，兩個連續的數的gcd是1，所以第一步是將原數列變成奇數數列或偶數數列，又因為對於長度n大於3時，偶數數列肯定要先出現大的gcd，所以第一步將原數列轉成偶數數列。

之後有趣的事情就出現了，可以發現，可以將形成的數列，奇數位上的數看“奇數數列”，偶數位上的數看成“偶數數列”，又重復第一步的過程。

在以上整個程中n都是大於3的，對於小於3的直接按“偶奇奇”的順序刪。

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<cstring>
 4 #include<string>
 5 #include<cmath>
 6 #include<algorithm>
 7 #include<queue>
 8 #include<stack>
 9 #include<deque>
10 #include<map>
11 #include<iostream>
12 using namespace std;
13 typedef long long  LL;
14 const double pi=acos(-1.0);
15 const double e=exp(1);
16 const int N = 100009;
17 
18 int con[1000009];
19 
20 int gcd(int a,int b)
21 {
22     int c;
23     while(b)
24     {
25         c=b;
26         b=a%b;
27         a=c;
28     }
29     return a;
30 }
31 
32 int main()
33 {
34     int i,p,j,n;
35     int cnt=0;
36     scanf("%d",&n);
37     for(i=1;i<=n;i++)
38         con[i]=i;
39     p=n;
40     while(p>0)
41     {
42         if(p==3)
43         {
44             printf("%d %d %d\n",gcd(gcd(con[1],con[2]),con[3]),gcd(con[2],con[3]),con[3]);
45             break;
46         }
47         else if(p>=2)
48         {
49             cnt=0;
50             int k=gcd(con[1],con[2]);
51             for(i=1;i<=p;i+=2)
52             {
53                 printf("%d ",k);
54                 if(i+1<=p)
55                     con[++cnt]=con[i+1];
56             }
57             p=cnt;
58         }
59         else if(p==1)
60         {
61             printf("%d\n",con[p]);
62             break;
63         }
64 
65     }
66     return 0;
67 }

CodeForces 1059C