Educational Codeforces Round 20 C 数学/贪心/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, ..., 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

6 3

Output

1 2 3

Input

8 2

Output

2 6

Input

5 3

Output

-1

题意：构造一个递增的长度为k 和为n 的数列，gcd尽可能的大.
题解：gcd 从1~i 所以寻找长度为k 首项为i 等差为i 数列和为n的一个序列
细细考虑 n应当模尽i 那么枚举n的因子 check 一下 取最大的i

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<set>
#include<vector>
#define ll __int64
using namespace std;
ll n,k;
bool fun(ll s){
    if(k*(k+1)*s/2>n)
        return false;
    if((n-k*(k-1)*s/2)%s==0)
        return true;
    return false;
}
int main()
{
    scanf("%I64d %I64d",&n,&k);
    if(k>(2*n/(1+k)))
        printf("-1
");
    else
    {
        ll e=1;
        for(ll i=1;i*i<=n;i++){
            if(n%i==0&&fun(i)) e=max(e,i);
            if(n%i==0&&fun(n/i)) e=max(e,n/i);
        }
        for(int j=1; j<k; j++)
            printf("%I64d ",j*e);
        printf("%I64d
",n-k*(k-1)*e/2);
    }
    return 0;
}