Split the Number（思维）

You are given an integer x. Your task is to split the number x into exactly n strictly positive integers such that the difference between the largest and smallest integer among them is as minimal as possible. Can you?

Input

The first line contains an integer T (1 ≤ T ≤ 100) specifying the number of test cases.

Each test case consists of a single line containing two integers x and n (1 ≤ x ≤ 10⁹, 1 ≤ n ≤ 1000), as described in the statement above.

Output

For each test case, print a single line containing n space-separated integers sorted in a non-decreasing order. If there is no answer, print  - 1.

Example

Input

1
5 3

Output

1 2 2

Note

The strictly positive integers are the set defined as: . That is, all the integers that are strictly greater than zero: .

题目意思：对于T组输入输出，给你一个n将其分为k份，要求分成的份最大值和最小值相差最小。

解题思路：这道题很简单，我看了队友写的代码后发现这个题还挺有意思的，这里给出解释。

例如10分成4份，用10直接去整除4得到的是2，我们可以这样理解平均每一份是ans 2，即n/k得到的是一种平均的分配，但是这样还没有全部分完n，还会有余数m=n%k，因为m必然是比k小的数，也就可以理解为有m个分数需要在原来的ans上再加一个上1，这样得到的分配能够保证最大值和最小值相差最小。

 #include<cstdio>
 #include<cstring>
 using namespace std;
 int main()
 {
     int t,ans,m,i,k,n;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d%d",&n,&k);
         if(n<k)
         {
             printf("-1\n");
             continue;
         }
         ans=n/k;
         m=n-ans*k;
         for(i=;i<k-m;i++)
         {
             printf("%d ",ans);
         }
         for(i=k-m;i<k;i++)
         {

             if(i==k-)
             {
                 printf("%d\n",ans+);
             }
             else
             {
                 printf("%d ",ans+);
             }
         }
     }
     return ;
 }