【Wunder Fund Round 2016 (Div 1 + Div 2 combined)A】【二进制拆分】Slime Combining 特殊规则二进制拆分

A. Slime Combining

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 n slimes. 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

#include<stdio.h>#include<iostream>#include<string.h>#include<string>#include<ctype.h>#include<math.h>#include<set>#include<map>#include<vector>#include<queue>#include<bitset>#include<algorithm>#include<time.h>using namespace std;void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }#define MS(x,y) memset(x,y,sizeof(x))#define MC(x,y) memcpy(x,y,sizeof(x))#define MP(x,y) make_pair(x,y)#define ls o<<1#define rs o<<1|1typedef long long LL;typedef unsigned long long UL;typedef unsigned int UI;template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }const int N = 0, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;int n;int s[32];void nice(){for (int i = 20; i >= 0; --i)if (n&(1 << i))printf("%d ", i+1);puts("");}int main(){while (~scanf("%d", &n)){nice(); continue;int top = 0;while (n){++top;if (n & 1)s[top] = top;else s[top] = 0;n >>= 1;}for (int i = top; i; --i)if (s[i])printf("%d ", s[i]);puts("");}return 0;}/*【trick&&吐槽】一定要看样例，搞错输出是必然要爆炸的。【题意】我们依次加入n个1，每次加在数列的最右侧。这个数列从右向左扫描，如果a[i]==a[i-1]，也就是出现两个权值相同的数，我们会把这2个数合并成1个权值大1的数。【类型】二进制拆分【分析】我们发现最后的数列很类似于二进制拆分数列。如果这个数二进制，从小到大(0bas)第i位是1，那么我们的答案就有i+1【时间复杂度&&优化】O(logn)*/