Codeforces H. Maximal GCD (greedy)

You are given positive integer number n. You should create such strictly increasing sequence of k positive numbers a₁, a₂, ..., a_k, that their sum is equal to n and greatest common divisor is maximal.

Greatest common divisor of sequence is maximum of such numbers that every element of sequence is divisible by them.

If there is no possible sequence then output -1.

Input

The first line consists of two numbers n and k (1 ≤ n, k ≤ 10¹⁰).

Output

If the answer exists then output k numbers — resulting sequence. Otherwise output -1. If there are multiple answers, print any of them.

Examples

Input

Copy

6 3

Output

Copy

1 2 3

Input

Copy

8 2

Output

Copy

2 6

Input

Copy

5 3

Output

Copy

-1

Ideas:

The problem is to give a number n, a number k, see if I can Couchu with a set of k elements of strictly increasing number of columns, and this set of numbers is n, but also let the greatest common divisor largest number of this group.

At first, a1 + a2 + ... + ak = n> = a1 + (a1 + 1) + (a1 + 2) + ... + (ak + k-1), solve for 1 <= a1 <= (nk * (k-1) / 2) / k, since the maximum divisor required, to find out a factor of all, from a large to a small specimens,

For example, to find a f a1 is a factor, if it wants to f gcd this set of numbers, it must also be f n factor, check, is not skipped, with regard to n / f, becomes to find a strictly increasing and the number of columns is n / f, would not have considered a common factor.

Beginning to put too much attention on a1, a1 put all possible values to go along, to find factors were, respectively, the result is wrong, and the program logic and chaos + complex, there are many special sentenced to place (I'm almost crazy) because each has a factor of 1 a1, a1 is the last for larger (from big to small checks) will always stop to 1, that to calculate the final result.

So then all of a factor all put in a set, but also easy to find the maximum, through all the factors from small to large. Enough. . . ?

The results timeout.

Think about it, how to do. Since last factor f if a1 are subject conditions are met, then it is also a factor n, then n factor to find out just fine ah. (9 (1> ◡ <1) 6), n is noted that the enumeration is not forget factor f to satisfy the constraints of the bounds of a1, it satisfies added to this set.

But, however, also noted that a start determination may be possible to construct such a series, a1 + a2 + ... + ak = n> = ka1 + k * (k-1) / 2> = k + k * (k- 1) / 2, that is over? NONONO, to be noted that the scope of the subject data, if to a 4 * 10 ^ 9 such, long long also save any ah, supposed that, with the double bar to keep k ...

Type	Size	Value range
No value type void	0 byte	No range
Boolean bool	1 byte	true false
Signed short short [int] / signed short [ int]	2 byte	-32768~32767
Unsigned short unsigned short [int]	2 byte	0~65535
Signed integer int / signed [int]	4 byte	-2147483648~2147483647
Unsigned int unsigned [int]	4 byte	0~4294967295
Signed long integer long [int] / signed long [ int]	4 byte	-2147483648~2147483647
Unsigned long unsigned long [int]	4 byte	0~4294967295
long long	8 byte	0~18446744073709552000
Signed char char / signed char	1 byte	-128~127
Unsigned char unsigned char	1 byte	0~255
Wide char wchar_t (unsigned short.)	2 byte	0~65535
Single-precision floating point type float	4 byte	-3.4E-38 3.4E + 38 ~
Double-precision floating-point double	8 byte	1.7E-308 ~ 1.7E + 308
long double	8 byte

(See table below blog on the source)

Code:

 1 #include <iostream>
 2 #include <cmath>
 3 #include <vector>
 4 #include <set>
 5 #include <algorithm>
 6 using namespace std;
 7 long long n;
 8 double k;
 9 set<long long> fac;
10 vector<long long> num;
11 int main()
12 {
13     cin >> n >> k;
14     if(k+k*(k-1)/2<=n)
15     {
16         long long uper1 = (n-k*(k-1)/2)/k;
17         long long lower1 = 1;
18         //cout << uper1 << " " << lower1 << endl;
19         long long qz = 1;
20         /*for(long long a = uper1; a>=lower1; a--)
21         {
22             for(long long i = 1; i<sqrt(a)+1; i++)
23             {
24                 if(a%i==0)
25                 {
26                     fac.insert(i);
27                     fac.insert(a/i);
28                 }
29             }
30         }*/
31         for(long long i = 1;i<sqrt(n)+1;i++)
32         {
33             if(n%i==0)
34             {
35                 if(i<=uper1)
36                 {
37                     fac.insert(i);
38                 }
39                 if(n/i<=uper1)
40                 {
41                     fac.insert(n/i);
42                 }
43             }
44         }
45         /*cout << "fac " << endl;
46         for(auto ite = fac.rbegin();ite!=fac.rend();ite++)
47         {
48             cout << *ite << " ";
49         }
50         cout << endl;*/
51         for(auto ite = fac.rbegin(); ite!=fac.rend(); ite++)
52         {
53             //cout << "ite " << *ite << endl;
54             int flag = 1;
55             num.clear();
56             long long sum = n/(*ite)-1;
57             int shu = 1;
58             num.push_back(shu);
59             if(sum)
60             {
61                 shu++;
62                 while(shu<=sum&&num.size()<k-1)
63                 {
64                     num.push_back(shu);
65                     sum -= shu;
66                     shu++;
67 
68                 }
69                 if(shu<=sum)
70                 {
71                     qz = *ite;
72                     //cout << "qz " << qz << endl;
73                     num.push_back(sum);
74                     break;
75                 }
76             }
77             else
78             {
79                 qz = *ite;
80                 break;
81             }
82         }
83         for(int i = 0;i<num.size();i++)
84         {
85             cout << qz*num[i] << " ";
86         }
87         cout << endl;
88     }
89     else
90     {
91         cout << -1 << endl;
92     }
93     return 0;
94 }

References:

Dazed and Confused 12138, C ++ basic data type size and scope of representation, https: //blog.csdn.net/weixin_40709898/article/details/79368310 (indirect reference, because the above blog is reprinted but could not find the original blog)