题干：

Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.

You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1.

You would like to see what the final state of the row is after you've added all nslimes. Please print the values of the slimes in the row from left to right.

Input

The first line of the input will contain a single integer, n (1 ≤ n ≤ 100 000).

Output

Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left.

Examples

Input

1

Output

1

Input

2

Output

2

Input

3

Output

2 1

Input

8

Output

4

Note

In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1.

In the second sample, we perform the following steps:

Initially we place a single slime in a row by itself. Thus, row is initially 1.

Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2.

In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1.

In the last sample, the steps look as follows:

1. 1
2. 2
3. 2 1
4. 3
5. 3 1
6. 3 2
7. 3 2 1
8. 4

题目大意：

朋友们送了一些史莱姆当作小N的生日礼物，总共有n个史莱姆，每个粘度为1。

小N打算用这些史莱姆玩游戏。他开始时先在桌子上放置一个史莱姆，然后逐个添加其他n-1个史莱姆。 

当添加史莱姆时，每次都将其放置在所有已放置的史莱姆的右侧。 一旦出现两个相邻的史莱姆粘度相同均为v ，它们就将组合在一起变为一个粘度v+1的史莱姆。

小N想知道在添加所有n个史莱姆之后，桌子上所有史莱姆的粘度是多少。 请从左到右打印史莱姆的粘度。

解题报告：

   不难发现规律是二进制中1的位置，然后乱搞就行了。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
const int MAX = 200 + 5;
int ans[MAX];
int tot;
int main()
{int n;cin>>n;int cur = 0;while(n) {cur++;if(n&1) ans[++tot] =  cur;n>>=1;}for(int i = tot; i>=1; i--) {printf("%d ",ans[i]);}return 0 ;
}