CF803C：Maximal GCD（思维 & 构造）

C. Maximal GCD

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given positive integer number n. You should create such strictly increasing sequence of k positive numbers a1, a2, ..., ak, 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

6 3

output

1 2 3

input

8 2

output

2 6

input

5 3

output

-1

题意：给出数字N和K，要求构造出K个严格递增的数字，是他们的和为N，且这K个数字的最大公因数最大。

思路：若这K个数字的gcd为d，那么N一定可以整除d，所以考虑从N的因子里选出这个d来，先求出K个数字可以表示的最小的数，即1+2+3+...+k，二分找出比这个数大的N的因子fact，那么N/fact就是这K个数的最大公因数，输出即可。

# include 
using namespace std;

typedef unsigned long long ull;
sets;
int main()
{
    ull n, k;
    scanf("%I64u%I64u",&n,&k);
    if(k+1 > 2*n*1.0/k) return 0*puts("-1");
    for(int i=1; i<=sqrt(n); ++i)
        if(n%i==0) s.insert(i),s.insert(n/i);
    ull max_num = (ull)(k*1.0/2*(k+1));
    auto it = s.lower_bound(max_num);
    ull sub = *it-max_num;
    max_num = n/(*it);
    for(int i=1; i