Friends and Cookies(思维)

Abood's birthday has come, and his n friends are aligned in a single line from 1 to n, waiting for their cookies, Abood has x cookies to give to his friends.

Here is an example to understand how Abood gives away the cookies. Suppose Abood has 4 friends and x cookies, then Abood will do the following:

1. Give a cookie to the 1^st friend.
2. Give a cookie to the 2^nd friend.
3. Give a cookie to the 3^rd friend.
4. Give a cookie to the 4^th friend.
5. Give a cookie to the 3^rd friend.
6. Give a cookie to the 2^nd friend.
7. Give a cookie to the 1^st friend.
8. Give a cookie to the 2^nd friend.
9. And so on until all the x cookies are given away.

Your task is to find how many cookies each friend will get. 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 ≤ 10¹⁸, 1 ≤ n ≤ 1000), in which x is the number of cookies Abood has, and n is the number of his friends.

Output

For each test case, print a single line containing n space-separated integers a₁, ..., a_n, in which a_i represents how many cookies the i^th friend got.

Example

Input

1
5 3

Output

2 2 1




解题思路：我将蛋糕的分配方式模拟一下
－　－　－　－　－　－　－
－　－　－　－　－　－
　　－　－　－　－　－　－
－　－　－　－　－　－　
　　－　－　－　－　－　－
　　　　　　－　－　－

我们可以看到的是：第一行每一个人都有，然后开始的奇数个前N-1个有，偶数个后N-1有，最后一行再判断是第奇数个还是偶数个就可以得出结果。

注意的是：没跑完第一行的情况需要特判一下,跑完第一行但是m<n也需要特判一下。

#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long int
using namespace std;
int main()
{
    int t;
    LL i,j,a[10010],m,n,x,y;
    scanf("%d",&t);
    while(t--)
    {
        memset(a,0,sizeof(a));
        scanf("%lld%lld",&m,&n);
        if(n==1)///特判一下当n为1的时候
        {
            printf("%lld\n",m);
        }
        else if(m<n)
        {
            for(i=0;i<m;i++)
            {
                a[i]=1;
            }
            for(i=0;i<n;i++)///只有前m个人有蛋糕，后面的人没有了蛋糕
            {
                if(i==n-1)
                {
                    printf("%lld\n",a[i]);
                }
                else
                {
                    printf("%lld ",a[i]);
                }
            }

        }
        else
        {
            for(i=0;i<n;i++)///第一行的分配
            {
                a[i]=1;
            }
            m=m-n;///分配完第一行之后剩下的蛋糕数量
            x=m/(n-1);///x表示剩下的行数
            y=m%(n-1);///y表示最后一行余下的蛋糕份数
            for(i=1;i<n-1;i++)
            {
                a[i]=a[i]+x;///除了第一个人和最后一个人其他的人再得到x份蛋糕
            }
            ///下面是对第一个人和最后一个人以及最后一行的蛋糕分配
            if(x%2==0)///行数为偶数
            {
                a[0]=a[0]+x/2;
                a[n-1]=a[n-1]+x/2;
                for(i=n-2;i>=n-2-y+1;i--)
                {
                    a[i]++;
                }
            }
            else///行数为奇数
            {
                a[0]=a[0]+x/2+1;
                a[n-1]=a[n-1]+x/2;
                for(i=1;i<=y;i++)
                {
                    a[i]=a[i]+1;
                }
            }
        }
        for(i=0;i<n;i++)///输出
        {
            if(i==n-1)
            {
                printf("%lld\n",a[i]);
            }
            else
            {
                printf("%lld ",a[i]);
            }
        }
    }
    return 0;
}