Educational Codeforces Round 20.C

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：输出一个k项数列使其满足：

1.各项之和为n  　　2.单增    　　3.最大公约数最大化。

思路：枚举n的因子q，找到满足n/q>=k*(k+1)/2 的最大q， 输出数列

PS：这题坑点挺多的。。比如数据范围  n,k 都是1e10 特判要考虑到这点。

代码：

 #include<stdio.h>
 #include<iostream>
 #include<algorithm>
 #include<string.h>
 #include<math.h>
 #include<stdlib.h>
 #include<ctype.h>
 #include<stack>
 #include<queue>
 #include<map>
 #include<set>
 #include<vector>
 #define ll long long
 #define  db double
 using namespace std;
 ;
 ;
 int main()
 {
     ll n,k;
     db s=sqrt(2.0);
     scanf("%lld%lld",&n,&k);
     ll ans=n;
     )/) {//必须要判断k的值，否则会溢出。
         puts("-1");
         ;
     }
     ll ma=;
     ;i<=sqrt(n);i++){
         &&n/i>=k*(k+)/){ //i,n/i都可能是最大因子
             ma=max<ll>(ma,i);
             )/) ma=max<ll>(ma,n/i);
         }
     }
     ;i<k;i++){
         printf("%lld ",i*ma);
         ans-=i*ma;
     }
     printf("%lld\n",ans);

 }