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
");
            continue;
        }
        ans=n/k;
        m=n-ans*k;
        for(i=0;i<k-m;i++)
        {
            printf("%d ",ans);
        }
        for(i=k-m;i<k;i++)
        {

            if(i==k-1)
            {
                printf("%d
",ans+1);
            }
            else
            {
                printf("%d ",ans+1);
            }
        }
    }
    return 0;
}