Phoenix has decided to become a scientist! He is currently investigating the growth of bacteria.
Initially, on day 1, there is one bacterium with mass 1.
Every day, some number of bacteria will split (possibly zero or all). When a bacterium of mass m splits, it becomes two bacteria of mass m2 each. For example, a bacterium of mass 3 can split into two bacteria of mass 1.5.
Also, every night, the mass of every bacteria will increase by one.
Phoenix is wondering if it is possible for the total mass of all the bacteria to be exactly n. If it is possible, he is interested in the way to obtain that mass using the minimum possible number of nights. Help him become the best scientist!
Input
The input consists of multiple test cases. The first line contains an integer t (1≤t≤1000) — the number of test cases.
The first line of each test case contains an integer n (2≤n≤109) — the sum of bacteria masses that Phoenix is interested in.
Output
For each test case, if there is no way for the bacteria to exactly achieve total mass n, print -1. Otherwise, print two lines.
The first line should contain an integer d — the minimum number of nights needed.
The next line should contain d integers, with the i-th integer representing the number of bacteria that should split on the i-th day.
If there are multiple solutions, print any.
Example
input
3
9
11
2
output
3
1 0 2
3
1 1 2
1
0
Note
In the first test case, the following process results in bacteria with total mass 9:
Day 1: The bacterium with mass 1 splits. There are now two bacteria with mass 0.5 each.
Night 1: All bacteria’s mass increases by one. There are now two bacteria with mass 1.5.
Day 2: None split.
Night 2: There are now two bacteria with mass 2.5.
Day 3: Both bacteria split. There are now four bacteria with mass 1.25.
Night 3: There are now four bacteria with mass 2.25.
The total mass is 2.25+2.25+2.25+2.25=9. It can be proved that 3 is the minimum number of nights needed. There are also other ways to obtain total mass 9 in 3 nights.
In the second test case, the following process results in bacteria with total mass 11:
Day 1: The bacterium with mass 1 splits. There are now two bacteria with mass 0.5.
Night 1: There are now two bacteria with mass 1.5.
Day 2: One bacterium splits. There are now three bacteria with masses 0.75, 0.75, and 1.5.
Night 2: There are now three bacteria with masses 1.75, 1.75, and 2.5.
Day 3: The bacteria with mass 1.75 and the bacteria with mass 2.5 split. There are now five bacteria with masses 0.875, 0.875, 1.25, 1.25, and 1.75.
Night 3: There are now five bacteria with masses 1.875, 1.875, 2.25, 2.25, and 2.75.
The total mass is 1.875+1.875+2.25+2.25+2.75=11. It can be proved that 3 is the minimum number of nights needed. There are also other ways to obtain total mass 11 in 3 nights.
In the third test case, the bacterium does not split on day 1, and then grows to mass 2 during night 1.
题目大意：
一开始有1个细菌，他的权值为1。白天的时候1个细菌可以分裂成2个，也可以不分裂，晚上的时候每个细菌会增加权值1，求最少要多少天能够到达所有权值的和恰好为n，并且给出对应n天每天有几个细菌要分裂。
题目分析：
我们来仔细分析一下题目：
一开始细菌的权值为1，有两种选择：
1）选择分裂，这样经过一天权值总和会加2，变为1+2
2）不分裂，这样经过一天权值总和会加1，变为1+1
因此，第一天变化范围为[2,1+2].
依次类推,第二天的变化范围为[3,1+2+4]，第三天的变化范围为[4,1+2+4+8]……
因为要求的是最小天数，所以，我们要每天分裂的细菌数尽可能最大。因此可以构造一个序列20,21,22,23…2x(序列的和sum<=n,x表示天数).这样就得到了细菌每天可以增长的权值的最大值。
1）如果这个序列的和sum==n，那么答案就正好为x.
2）如果sum!=n，那么就要在序列中加一个n-sum，那么答案即为x+1.
再将加上n-sum后的序列进行排序，第i天细菌的分裂数即为：f[i]-f[i-1]。
ac代码：
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <map>
#include <queue>
#include <vector>
#include <algorithm>
#include <iomanip>
#define LL long long
using namespace std;
const int N=1e5+5;
int main()
{
int t;
cin>>t;
while(t--)
{
int n;
cin>>n;
vector<int> f; //f存储权值每天的增加值
for(int i=1;i<=n;i<<=1) //按上述方法构造f
{
f.push_back(i);
n-=i;
}
if(n) f.push_back(n); //放入n-sum
sort(f.begin(),f.end()); //排序
cout<<f.size()-1<<endl; //输出答案即可
for(int i=1;i<f.size();i++)
{
cout<<f[i]-f[i-1]<<' ';
}
cout<<endl;
}
return 0;
}