CodeForces 618A-Slime Combining【模拟】-蒲公英云

CodeForces 618A-Slime Combining【模拟】

你的名字 2022-08-21 04:15 99阅读 0赞

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 othern - 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 valuev, you combine them together to create a slime with valuev + 1.

You would like to see what the final state of the row is after you’ve added alln 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, wherek is the number of slimes in the row after you’ve finished the procedure described in the problem statement. Thei-th of these numbers should be the value of thei-th slime from the left.

Examples

Input

Output

Input

Output

Input

Output

2 1

Input

Output

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 value2. 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
2
2 1
3
3 1
3 2
3 2 1

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
    int n;
    while(~scanf("%d",&n)){
        /*if(n==1)
        {
            printf("1\n");
            continue;
        }
        if(n==2)
        {
            printf("2\n");
            continue;
        }*/
        int ji;
        while(n)
        {
            int w=0;
            while(pow(2,w)<=n)
            {
                w++;
            }
            printf("%d ",w);
            w--;
            n=n-pow(2,w);
        } 
        printf("\n");
    }
    return 0;
}