Maximal GCD

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 
            Copy 
          
6 3
 output 
            Copy 
          
1 2 3
 input 
            Copy 
          
8 2
 output 
            Copy 
          
2 6
 input 
            Copy 
          
5 3
 output 
            Copy 
          
-1

The idea of this question is to arrange the common divisors of n from large to small, first to find the larger common divisor m1, and then find another m2 as the corresponding factor. If allocating 1, 2, 3, ..., k m1 to k numbers are enough points, then m1 is a feasible greatest common divisor.

The key point is the special judgment, which is the case when n cannot satisfy the condition even if 1 is used as a common divisor. When it is necessary to judge that k is large to a certain value, n will exceed 1e10 to meet the condition, and if you continue to judge k*(k+1)/2, there will be a situation where this value exceeds the long long type, so this condition must be to add.

#include <cstdio>
#include <algorithm>
#include<iostream>
#define ll long long
using namespace std;
bool cmp(ll a,ll b)
{
    return a > b;
}
int a[100055];
intmain()
{
    ll n,k,m1,m2,num=0,last;
    scanf("%I64d%I64d",&n,&k);
    if(n<(k+1)*k/2||k>1e6) {printf("-1\n");return 0;}
    ll i;
    for(i=1;i*i<n;i++)
        if(n%i==0) a[num++]=i,a[num++]=n/i;
    if(n%i==0) a[num++]=i;
    sort(a,a+num,cmp);
    for(i=0;i<num;i++)
    {
        m1=a[i];
        m2=n/a[i];
        if(m2>=(k+1)*k/2) break;
    }
    last=m2-k*(k-1)/2;
    for(i=1;i<k;i++)
        printf("%I64d ",i*m1);
    printf("%I64d\n",last*m1);
    return 0;
}

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